Lecture 36. Linearization Near Critical Points1. A Caveat, Or A Term's A Term No Matter How SmallDifferential EquationsLECTURE 36Linearization Near Critical PointsLast lecture, we began considering the nonlinear systemx0= f(x, y) (36.1)y0= g(x, y).We saw that near the critical points (points (x0, y0) satisfying f (x0, y0) = g(x0, y0) = 0), whichcorrespond to equilibrium solutions of the system, we can approximate the nonlinear system (36.1)by the linear systemx0y0= ∂f∂x(x0, y0)∂f∂y(x0, y0)∂g∂x(x0, y0)∂g∂y(x0, y0)!xy. (36.2)We can then analyze these linear systems to conclude what we can about the behavior ofsolutions near the equilibria. Notice that this is entirely a local method: we don’t obtain informationabout what happens far away from the critical points. However, we can sometimes use commonsense to determine other features of the phase portrait.Example 36.1. Find the critical points of the following systems, then determine the lineariza-tions near them.(i) x0= y2− 3x + 2 y0= x2− y2The first task is to find the critical points. We do this by setting x0= f(x, y) andy0= g(x, y) and then looking for the values of x and y that make both f and g zero.Looking at the two equations, we might decide that it’s easy to conclude something fromg(x, y) = x2− y2= 0. Indeed, we see that we must havex2= y2.Thus, the equation f(x, y) = y2− 3x + 2 = 0 becomesx2− 3x + 2 = 0(x − 2)(x − 1) = 0.So we must have x = 2 or x = 1. Now, we know x2= y2. So if x = 2, y2= 4, and so y = ±2.Similarly, if x = 1, y2= 1, and so we get y = ±1.Summarizing, we have four critical points: (1, 1), (1, −1), (2, 2), (2, −2).Now, let’s find the linearizations of the system near each of these points. The Jacobianmatrix of the nonlinear system at a general point (x, y) isfxfygxgy=−3 2y2x −2y.1Differential Equations Lecture 36: Linearization Near Critical PointsTo get the coefficient matrix of our linearized system near one of our critical points, all wehave to do is evaluate these terms at the point. Thus we get the following linearizations.(1, 1) : x0=−3 22 −2x(−1, −1) : x0=−3 −22 2x(2, 2) : x0=−3 44 −4x(2, −2) : x0=−3 44 4x(ii) x0= y y0= −x + x3Again, the first task is to find the critical points. For x0= 0, we need to require thaty = 0. Then, for y0= 0, we need x = x3. This means x = 0, ±1. So our critical points are(0, 0), (−1, 0), and (1, 0).The Jacobian matrix of the system at (x, y) is0 1−1 + 3x20,so we end up with the following linearizations at each of the critical points.(0, 0) : x0=0 1−1 0x(−1, 0) : x0=0 12 0x(1, 0) : x0=0 12 0x So, we’ve found the critical points and the linearizations of the original nonlinear system nearthem. Now what? Normally, we would go through our linear systems analysis to determine the typeand stability of each critical point: remember, we can do this from just determining the eigenvalues.Ideally, the type and stability of the origin in these linear systems would correspond to the typeand stability of the associated critical point in the nonlinear system.Exercise. For each of the critical points in both systems given in Example 36.1, determinethe type and stability of the origin in the linearized systems.However, there’s...1. A Caveat, Or A Term’s A Term No Matter How SmallIn turning our nonlinear system near a critical point into the linear system 36.2, we disregardedterms of order two or higher on the grounds that they were very small. Is it really safe to do that?In principle, after all, those terms could influence the behavior of the system in ways that makethis linearization method unreliable. It turns out that as long as linearized system’s critical pointis not one of several ”borderline” cases, this doesn’t affect the qualitative type of the critical point.2Differential Equations Lecture 36: Linearization Near Critical Points1.1. Simple Eigenvalues. If the linearized system near (x0, y0) has simple (i.e., distinct)eigenvalues, the only problems occur when the linearization predicts we will have a center. Spirals,nodes, and saddles are all preserved. Moreover, stability is preserved: we can’t go from having anunstable spiral in the linearized system to having a stable spiral in the nonlinear system. Thus ifour linearization predicts we have a spiral, node, or saddle, we can conclude that the critical pointwill be of the same type.Here’s an example of how this fails when the linearization predicts a center.Example 36.2. Show that the two systemsx0= −y + x(x2+ y2) y0= x + y(x2+ y2)x0= −y − x(x2+ y2) y0= x − y(x2+ y2)have the same linearized systems at the critical point (0, 0), but different phase portraits.The Jacobian matrix of the first system is3x2+ y2−1 + 2xy1 + 2xy x2+ 3y2.The Jacobian matrix of the second system is−3x2− y2−1 − 2xy1 − 2xy −x2− 3y2.At (0, 0), then, both have the same linearization, namelyx0=0 −11 0x.The eigenvalues of this matrix are ±i, and so this linearized system has a center.However, we can analyze this particular system more directly by using polar coordinates. If wedo that, the two systems becomer0= r3θ0= 1r0= −r3θ0= 1.Thus, non-zero trajectories for the first system will expand, as r0> 0 for all r > 0, whilenon-zero trajectories for the second system will decay as r0< 0. We actually have an unstablespiral for the first system and an asymptotically stable spiral for the second system, even thoughthe linearization predicted a center. These spirals, however, rotate in the same direction as thepredicted center. The previous example demonstrated the sort of thing we would need to do if we actually wantedto determine the phase portrait near a predicted center. In general, this will require a more directcomputation rather than a linearization. For our purposes, though, it will generally suffice to beaware that we can’t trust the prediction of a center; even small higher order terms can throw acenter to a spiral of either stability.1.2. Repeated Eigenvalues. In the case of a repeated eigenvalue λ, we ended up gettingeither a star node or a degenerate node depending on whether λ was complete or defective. Thisis also a delicate case; we will, however, be able to conclude something.If lambda is complete, and the linearization predicts a star node, we can conclude only thatthe nonlinear system will have a node (possibly degenerate or star) at
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