PAGE 1

7.3 The Eigenvalue Method (Continued)

\[ \vec{x}' = A \vec{x} \]

If \( A \) is a constant matrix, \( n \times n \), then there are \( n \) eigenvalue/eigenvector pairs:

\[ \lambda_1, \lambda_2, \dots, \lambda_n \]

\[ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \]

\( n \) solutions: \( e^{\lambda_i t} \vec{v}_i \)

General solution: linear combo of them

\[ \vec{x}(t) = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 + \dots + c_n e^{\lambda_n t} \vec{v}_n \]

Review and Today's Topic

Last time: real distinct \( \lambda \)'s
Today: complex \( \lambda \)'s

PAGE 2

Let's start w/ a slope field

\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \vec{x} \]

Figure: A slope field on a 2D coordinate system with axes x1 and x2 showing a clockwise rotational flow.

\[ \vec{x} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies \vec{x}' = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

\[ \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \implies \vec{x}' = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \]

\[ \vec{x} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \implies \vec{x}' = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

\( \vdots \)

PAGE 3

Spiral Direction Field and Complex Eigenvalues

Figure: A vector field plot on a Cartesian plane with axes from -1 to 1, showing a green spiral trajectory moving outwards.

complex eigenvalues

↓

spiral direction field

PAGE 4

Solving Systems with Complex Eigenvalues

\[ \vec{x}' = \underbrace{\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}}_{A} \vec{x} \quad \text{solutions are still } e^{\lambda t} \vec{v}_i \]

Characteristic Equation

\[ \det(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 1 \\ -1 & 1 - \lambda \end{vmatrix} = 0 \]

\[ (1 - \lambda)^2 + 1 = 0 \]

\[ (1 - \lambda)^2 = -1 \]

\[ 1 - \lambda = \pm i \]

\[ \lambda = 1 + i, \quad 1 - i \]

always conjugate pairs

Eigenvectors

\[ \lambda = 1 + i \]

\[ (A - \lambda I) \vec{v} = \vec{0} \]

\[ \begin{bmatrix} -i & 1 & : & 0 \\ -1 & -i & : & 0 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 1 & i & : & 0 \\ 0 & 0 & : & 0 \end{bmatrix} \]

\[ \vec{v} = \begin{bmatrix} 1 \\ i \end{bmatrix} \]

PAGE 5

Complex Eigenvalues and Solutions

\(\lambda = 1 - i\)

\[\begin{bmatrix} i & 1 & \vdots & 0 \\ -1 & i & \vdots & 0 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 1 & -i & \vdots & 0 \\ 0 & 0 & \vdots & 0 \end{bmatrix}\]

\(\vec{v} = \begin{bmatrix} 1 \\ -i \end{bmatrix}\)

\(\vec{v}\)'s are also in conjugate pairs

Two Solutions in the Form \(e^{\lambda t} \vec{v}\)

For \(\lambda = 1 + i\), \(\vec{v} = \begin{bmatrix} 1 \\ i \end{bmatrix}\):

\[e^{\lambda t} \vec{v} = e^{(1+i)t} \begin{bmatrix} 1 \\ i \end{bmatrix}\]

Euler's formula: \(e^{it} = \cos t + i \sin t\)

\[= e^t e^{it} \begin{bmatrix} 1 \\ i \end{bmatrix} = e^t (\cos t + i \sin t) \begin{bmatrix} 1 \\ i \end{bmatrix}\]\[= e^t \begin{bmatrix} \cos t + i \sin t \\ -\sin t + i \cos t \end{bmatrix} = \underbrace{e^t \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix}}_{\text{real part}} \oplus i \underbrace{e^t \begin{bmatrix} \sin t \\ \cos t \end{bmatrix}}_{\text{imaginary part}}\]

PAGE 6

Repeat for \(\lambda = 1 - i\), \(\vec{v} = \begin{bmatrix} 1 \\ -i \end{bmatrix}\):

\[e^{(1-i)t} \begin{bmatrix} 1 \\ -i \end{bmatrix} = e^t e^{i(-t)} \begin{bmatrix} 1 \\ -i \end{bmatrix}\]\[= e^t (\cos t - i \sin t) \begin{bmatrix} 1 \\ -i \end{bmatrix}\]\[= e^t \begin{bmatrix} \cos t - i \sin t \\ -\sin t - i \cos t \end{bmatrix}\]\[= \underbrace{e^t \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix}}_{\text{real part}} \ominus i \underbrace{e^t \begin{bmatrix} \sin t \\ \cos t \end{bmatrix}}_{\text{imaginary part}}\]

Again, complex conjugate pairs.

Complex \(\lambda\)'s: one solution is \(\vec{u} + i\vec{v}\) and the other is \(\vec{u} - i\vec{v}\)

PAGE 7

Fundamental Solutions for General Solutions

Just like with scalar equations, we use the real and imaginary parts of either solution as the fundamental solutions for the general solution.

One solution: \( \vec{u} + i\vec{v} \)
General solution: \( c_1 \vec{u} + c_2 \vec{v} \) (no \( i \))

Here, the general solution to \( \vec{x}' = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \vec{x} \) is

\[ \vec{x}(t) = c_1 e^t \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix} + c_2 e^t \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} \]

Complex \( \lambda \)'s \( \rightarrow \) spirals \( \rightarrow \) cosines + sines

PAGE 8

Choosing Eigenvectors

There is a complication: how we choose eigenvector

Example

\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \vec{x} \]

\( \lambda = \pm i \)

Work out one solution, identify real and imaginary parts

Let's choose \( \lambda = -i \)

Solve \( (A - \lambda I) \vec{v} = 0 \)

\[ \begin{bmatrix} i & 1 & : & 0 \\ -1 & i & : & 0 \end{bmatrix} \rightarrow \dots \rightarrow \begin{bmatrix} 1 & -i & : & 0 \\ 0 & 0 & : & 0 \end{bmatrix} \]

\[ \vec{v} = \begin{bmatrix} a \\ b \end{bmatrix} \]

\( b \) is free: \( b = r \), \( a = ib = ir \)

\[ \vec{v} = \begin{bmatrix} ir \\ r \end{bmatrix} = r \begin{bmatrix} i \\ 1 \end{bmatrix} \]

choose ANY \( r \neq 0 \)

PAGE 9

If \( r = -i \), \( \vec{v} = \begin{bmatrix} 1 \\ -i \end{bmatrix} \)

If \( r = i \), \( \vec{v} = \begin{bmatrix} 1 \\ i \end{bmatrix} \)

Working out the solution

Let's work out the solution using these choices.

Start with \( \lambda = -i \), \( \vec{v} = \begin{bmatrix} 1 \\ -i \end{bmatrix} \)

\[ e^{\lambda t} \vec{v} = e^{-it} \begin{bmatrix} 1 \\ -i \end{bmatrix} = (\cos t - i \sin t) \begin{bmatrix} 1 \\ -i \end{bmatrix} \]\[ = \begin{bmatrix} \cos t - i \sin t \\ -\sin t - i \cos t \end{bmatrix} = \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix} - i \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} \]

The first vector is labeled \( \vec{u} \) and the second vector is labeled \( \vec{v} \).

General solution:

\[ \vec{x} = c_1 \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix} + c_2 \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} \]

PAGE 10

Repeat with \( \lambda = -i \), \( \vec{v} = \begin{bmatrix} i \\ 1 \end{bmatrix} \)

\[ e^{\lambda t} \vec{v} = e^{-it} \begin{bmatrix} i \\ 1 \end{bmatrix} \]\[ = (\cos t - i \sin t) \begin{bmatrix} i \\ 1 \end{bmatrix} \]\[ = \begin{bmatrix} i \cos t + \sin t \\ \cos t - i \sin t \end{bmatrix} \]\[ = \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} + i \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix} \]

The first vector is labeled \( \vec{u} \) and the second vector is labeled \( \vec{v} \).

General solution:

\[ \vec{x} = c_1 \begin{bmatrix} \sin t \\ \cos t \end{bmatrix} + c_2 \begin{bmatrix} \cos t \\ -\sin t \end{bmatrix} \]

These two solutions are the same even though they look different.

PAGE 11

Example

\[ \vec{x}' = \begin{bmatrix} 3 & 0 & 1 \\ 9 & -1 & 2 \\ -9 & 4 & -1 \end{bmatrix} \vec{x} \]

Possibilities:

3 real \(\lambda\)'s (distinct or repeated) (next time)
2 complex and one real

\[ \lambda = 3, -1+i, -1-i \] \[ \vec{v} = \begin{bmatrix} 4 \\ 9 \\ 0 \end{bmatrix}, \begin{bmatrix} -4-i \\ -9+2i \\ 17 \end{bmatrix}, \begin{bmatrix} -4+i \\ -9-2i \\ 17 \end{bmatrix} \]

General solution is formed the same way:

\[ e^{3t} \begin{bmatrix} 4 \\ 9 \\ 0 \end{bmatrix} \rightarrow \lambda = 3, \vec{v} = \begin{bmatrix} 4 \\ 9 \\ 0 \end{bmatrix} \]

Then work out either

\[ e^{(-1+i)t} \begin{bmatrix} -4-i \\ -9+2i \\ 17 \end{bmatrix} \]

or

\[ e^{(-1-i)t} \begin{bmatrix} -4+i \\ -9-2i \\ 17 \end{bmatrix} \]

and get the real and imaginary parts.