I began by reminding the class of the eigenvalue story, using as an example the settling-tank system
x' = -3x + y + 2r
y' = 3x - 3y
from Class 22. In general, one solves an inhomogeneous linear system (*) in a two-step process:
(1) Find n linearly independent solutions to the homogeneous equation, u1, ..., un, where n is the number of variables.
(2) Find a particular solution up to (*).
Then the general solution to (*) is up + a1 u1 + ... + an un.
For (1), begin by writing down the characteristic polynomial and finding its eigenvalues: lambda^2 + 6 lambda + 6 in this case, with roots lambda1 = -3 + sqrt 3, lambda2 = -3 - sqrt 3. Already you learn that all solutions of (H) decay to 0 (since the real parts are negative) with out oscillation (since the eigenvalues are real). (On Wednesday I said you'd get oscillation but I was wrong.)
Then you find the corresonding eigenvectors, [1;-sqrt 3] and [1;sqrt 3]. Thus
u1 = e^{(-3+sqrt 3)t}[1;-sqrt 3], u2 = e^{(-3-sqrt 3)t}[1; sqrt 3].
For (2), try a constant solution (by analogy with the method of undetermined coefficients). Then x' = y' = 0, so we wish to solve
0 = -3x + y + 2r 0 = 3x - 3y
I have to assume r constant; then x = y = r, so yp = [r;r], and the general solution is
u = [r;r] + a e^{(-3+sqrt 3)t}[1;-sqrt 3] + b e^{(-3-sqrt 3)t}[1; sqrt 3].
The first term is the steady state; the second two make up a transient.
Complex roots: For example take A = [0 1; -10 -2]. The characteristic polynomial is conveniently computed using the general fact that the characteristic polynomial of A = [a b; c d] is
lambda^2 - (tr A)lambda + (det A); tr A = a + d, det A = ad - bc.
We get lambda^2 + 2 lambda + 10, with roots lambda1 = -1 + 3i, lambda2 = -1 - 3i.
You learn that all solutions decay to 0 but now exhibit oscillation.
Swallowing our fear, subtract lambda1 from the diagonal of A and find an eigenvector, [1; -1 + 3i]. An eigenvector for lambda2, which is the conjugate of lambda1, is gotten by conjugating: [1, -1 - 3i]. We get two normal modes:
e^{(-1+3i)t}[1;-1+3i], e^{(-1-3i)t}[1;-1-3i].
To get real solutions we can take approprate linear combinations of these: half the sum, and the difference divided by 2i, i.e., the real and imaginary parts of the first. To find them write it as
u = e^{-t}(cos(3t) + i sin(3t))([1;-1]+i[0;3]) so
u1 = Re u = e^{-t}(cos(3t)[1;-1] - sin(3t)[0;3])
u2 = Im u = e^{-t}(cos(3t)[0;3] + sin(3t)[1;-1]).
This particular matrix A is the companion matrix of the second order ODE x" + 2x' + 10x = 0; x' = y, y' = -10x - 2y. The upper entries in u1 and u2 are independent solutions to x" + 2x' + 10x = 0, and the lower entries are the derivatives of those solutions. Note also that the characteristic polynomial of the second order equation is the same as the characteristic polynomial of the companion matrix.
Repeated roots: For example take A = [1 1; -1 3]. The characterstic polynomial has roots 2, 2. There is only one normal mode (up to scalar multiple). The eigenvalue is [1;1], so e^{2t}[1;1] is a solution. There is another of the form e^{2t}(t[1;1] + b), where b is some vector which we have to solve for. You can, and you get b = [0;2] (or b = [0;2] + a[1;1] for any a). This is the "incomplete case."
Some Part I problems: EP 5.4 # 1, 5, 13, 17.