Notes (Ch 9): Poincare Classification - 3 Lecture Note

Notes (Ch 9): Poincare Classification • Earlier, the autonomous DE was: y 0 = f (y). Goal: Find and classify the equilibria (we looked at plot of y v. y 0 ). • Goal today: Classify the origin in x0 = Ax Summary from last time Solve x = Ax: Find the eigenvalues/eigenvectors (through the trace, det and discriminant) √ Tr(A) ± ∆ λ2 − Tr(A)λ + det(A) = 0 λ= 2 Depends on ∆: • Real λ1 , λ2 give two eigenvectors, v1 , v2 : x(t) = C1 eλ1 t v1 + C2 eλ2 t v2 • Complex λ = a + ib, v (we only need one):

x(t) = C1 Real eλt v + C2 Imag eλt v • One eigenvalue, one eigenvector v. Get w that solves (A − λI)w = v. Then: x(t) = eλt (C1 v + C2 (tv + w)) So the type of solution depends on ∆, and in particular, where ∆ = 0: ∆=0 ⇒ 0 = (Tr(A))2 − 4det(A) This is a parabola in the (Tr(A), det(A)) coordinate system. This cuts the plane into several regions, which we examine below (HANDOUT: Poincaré Diagram) 1 Template Examples Use the Poincaré diagram to locate these. The diagram gives “template” shapes, not the actual shapes (we’ll see examples below). • Two distinct eigenvectors: Saddle, source or sink. – Both negative eigenvalues: Sink – Both positive eigenvalues: Source – Mixed sign: Saddle • Complex eigenvalues: Centers and Spirals – Real(λ) = 0 Center (Purely periodic, closed curves) – Real(λ) > 0: Spiral Source – Real(λ) < 0: Spiral Center • One eigenvector: We have a degenerate source (positive λ) or sink (negative λ). Sometimes hard to distinguish these graphically from a spiral. Using the Poincaré Diagram In the examples on the next few pages, if the matrix had a double eigenvalue with only one eigenvector, then the generalized eigenvector is given (along with the regular eigenvector). 2 Figure 1: First set of four phase planes. From top to bottom, Saddle, Degenerate Sink, Saddle, Line of Stable fixed points. First set of examples (See Figure ) System     3 0 0 −2 −1 0 0 −1 Tr(A) det(A) ∆ Poincare λ  1 −6 25 Saddle 0 Degen Sink  −2 2 −1 3 −2  4 −3 8 −6  0 −2 1 −1 0 4 Saddle 4 Line of Stable 3 V  1 0 0 1   1 0 0 1   1 1 3 1   1 3 2 4  3, −2 −1, −1 −1, 1 −2, 0 Figure 2: Second set of four phase planes. From top to bottom, Center, Degenerate Source, Spiral Sink, Sink. System   1 −1 1 3 ∆ Poincare λ  1 2 −5 −1   Tr(A) det(A)  0 9 −36 Center 0 Degen Source −4 Spiral Sink  4 − 12 1 −1 − 12  −1 −1 0 − 41  V −1 4 5/4 3i  2, 2 1/4 9/16 4 Sink − 12 + i −1, −1/4 −1 0 1 1   −5/4 2 −1 + 3i 1 i    1 −4 0 3  Figure 3: Third set of four phase planes. From top to bottom, Spiral Source, Saddle, Center, “Uniform Motion”. System     Tr(A) det(A) 3 −2 4 −1  2 −1 3 −2  0 2 −2 0  4 −2 8 −4  2 0 5 −1 ∆ Poincare −16 Spiral source 4 Saddle λ V  1 1−i   1 1 3 1  1 + 2i −1, 1  0 0 9 0 −36 Center 0 Uniform Motion 5 2i  0, 0 −i 1  1 0 2 −1/2  Homework: 1. Give the solution to the four systems x0 = Ax from the previous page (page 5). 2. For the three systems below x0 = Ax, try to give a rough sketch of the behavior of the solutions near the equilibrium (the origin):      1  −1 0 1 −1 −2 1 A= A= A= 0 −1 1 3 −1 − 12 3. For the systems below, use the Poincaré diagram to classify the equilibrium solution. If necessary, first convert the equation to a system of first order equations:   5 −1 0 (a) x = x 3 1 (b) y 00 + y 0 + 3y = 0 (c) 2y 00 − 3y 0 + 4y = 0   3 −4 0 (d) x = x 1 −1 4. For the following nonlinear systems, find the equilibrium solutions (the derivatives are with respect to t, as usual). (a) x0 = x − xy, y 0 = y + 2xy (b) x0 = y(2 − x − y), y 0 = −x − y − 2xy 5. Explain how the classification of the origin changes by changing the α in the system:   0 α 0 x = x 1 −2 6