Math 33B - Final Exam Review Sheet

I. Linear systems of differential equations: (See Sections 9.1 and 9.2)

y = Ay where A is an n n matrix with constant entries General idea: Find n linearly independent solutions y1, . . . ,yn. The general solution is

then y = C1y1 + . . .+ Cnyn.

To find solutions, find the eigenvalues of A by solving det(A I) = 0. Then for eacheigenvalue , find its corresponding eigenvectors by solving (A I)v = 0. For a distinct eigenvalue and its corresponding eigenvector v, form the solutiony = etv.

For a pair of complex conjugate eigenvalues = a bi, pick one and compute itseigenvector v, then form the solution y = etv as before, and simplify using Eulersformula. Then the real part of this gives one solution, and the imaginary part givesanother.

For a repeated eigenvalue with two linearly independent eigenvectors v1, v2, justform the two solutions y1 = e

tv1, y2 = etv2.

For a repeated eigenvalue with only one eigenvector v (up to linearly indepen-dence), find a generalized eigenvector w by solving (A I)w = v. Then form thesolution y = et(w + vt).

II. Classification of the Equilibria of y = Ay where A is a 2 2 matrix with trace T anddeterminant D (See Sections 9.3 and 9.4)Note: In all cases except the last, there is a single equilibrium point, and it is at the origin.Generic Cases:

Real eigenvalues, one positive, one negative (D < 0): Saddle Distinct real e-values, both negative (D > 0, T 2 4D > 0, T < 0): Nodal Sink Distinct real e-values, both positive (D > 0, T 2 4D > 0, T > 0): Nodal Source Complex e-values with negative real part (T 2 4D < 0, T < 0): Spiral Sink Complex e-values with positive real part (T 2 4D < 0, T > 0): Spiral Source

Degenerate Cases:

Purely imaginary eigenvalues (D > 0, T = 0): Center Negative e-value of alg. multiplicity two (T 2 4D = 0, T < 0): Degenerate Sink

(Like a nodal sink, but a solution may curve significantly as it approaches the origin, likeits almost a spiral. Or no curving at all: a star sink.)

Positive e-value of alg. multiplicity two (T 2 4D = 0, T > 0): Degenerate Source(Like a nodal source, but a solution may curve significantly as it leaves the origin, likeits almost a spiral. Or no curving at all: a star source.)

One of the eigenvalues is 0 (D = 0): No Catchy NameIn this case there are infinitely many equilibria, arranged along a line in the plane. IfT < 0 they are all stable. If T > 0, they are all unstable. If T = 0 they are neither. Allother solutions are straight lines.

III. Fundamental Matrices (See Section 9.9)

Definition: A fundamental matrix for A(t) is a matrix Y (t) that is invertible andsatisfies Y (t) = A(t)Y (t).

To compute a fundamental matrix for A(t), find n linearly independent solutions of then n linear system y(t) = A(t)y(t), and create a matrix with those solutions as itscolumns.

Given a fundamental matrix Y (t) for A(t), if M is any invertible constant matrix, thenY (t)M is also a fundamental matrix for A(t), and furthermore all fundamental matricesfor A(t) arise in this way.

IV. Variation of Parameters for Inhomogeneous Linear Systems (See Section 9.9)To solve y(t) = A(t)y(t)+g(t), first find a fundamental matrix Y (t) for A(t). Then a solutionis given by

y(t) = Y (t)

Y (t)1g(t)dt.

V. The Matrix Exponential eAt (See Section 9.9, and maybe 9.6)

Definition: eAt = n=0 tnn!An = I +At+ 12A2t2 + 16A3t3 + . . .. eAt is the unique fundamental matrix Y (t) for A satisfying Y (0) = I.

Similarly, eA(tt0) is the unique fundamental matrix Y (t) for A satisfying Y (t0) = I. If Y (t) is any fundamental matrix for A, then eAt = Y (t)Y (0)1.

Likewise, if Y (t) is any fundamental matrix for A, then eA(tt0) = Y (t)Y (t0)1. The solution of the initial value problem y = Ay, y(t0) = y0 is

y = eA(tt0)y0.

VI. Qualitative Analysis of Autonomous Nonlinear Systems (See Sections 10.1 and 10.3){x = f(x, y)y = g(x, y)

The x-nullcline is the set of points (x, y) where f(x, y) = 0, i.e., where the arrows allpoint straight up or down. The y-nullcline is the set of points (x, y) where g(x, y) = 0,i.e., where the arrows all point straight left or right.

A point (x0, y0) is an equilibrium if f(x0, y0) = 0 and g(x0, y0) = 0, i.e., if it is on bothnullclines.

The Jacobian of the system is the matrix

J =

(fx

fy

gx

gy

).

The linearization of the system at an equilibrium (x0, y0) is the homogeneous linearsystem y = J(x0, y0)y.

If the linearization of the system at (x0, y0) is one of the five generic types (saddle, nodalsink, nodal source, spiral sink, spiral source), then the equilibrium at (x0, y0) is of thesame type.