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7.6 Complex Eigenvalues

\[ \vec{x}' = A\vec{x} \quad \text{where } A \text{ has complex eigenvalues} \]

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

\[ \begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2 + 1 = 0 \implies \lambda = \pm i \]

Look at the direction field to get some clue about the solution

Figure: A 2D direction field on x1 and x2 axes showing vectors forming a counter-clockwise circular flow around the origin.

If \( \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), then \( \vec{x}' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

If \( \vec{x} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \), then \( \vec{x}' = \begin{bmatrix} -1 \\ 0 \end{bmatrix} \)

The direction field shows the solutions are circles (ovals):

Each component of the solution is periodic
Suggesting sines and cosines

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\[ \text{Solution: } e^{\lambda t} \vec{v} \]

Where \( \lambda \) is the eigenvalue and \( \vec{v} \) is the corresponding eigenvector.

Eigenvector for \( \lambda = i \)

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

\[ \begin{bmatrix} -i & -1 & \big| & 0 \\ 1 & -i & \big| & 0 \end{bmatrix} \xrightarrow{iR_2 + R_1} \begin{bmatrix} 0 & 0 & \big| & 0 \\ 1 & -i & \big| & 0 \end{bmatrix} \]

\[ \vec{v} = \begin{bmatrix} i \\ 1 \end{bmatrix} \quad \text{eigenvector is also complex} \]

Eigenvector for \( \lambda = -i \)

Repeat the procedure above:

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

Eigenvalue/Eigenvector Pairs:

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

\[ \lambda = -i, \quad \vec{v} = \begin{bmatrix} -i \\ 1 \end{bmatrix} \]

These are always complex conjugate pairs.

→ We can never have an odd number of complex eigenvalues.

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Complex Solutions to Systems of Differential Equations

Solution 1

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

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

Solution 2

\[ \begin{aligned} \vec{x}^{(2)} &= e^{-it} \begin{bmatrix} -i \\ 1 \end{bmatrix} \\ &= (\cos(-t) + i \sin(-t)) \begin{bmatrix} -i \\ 1 \end{bmatrix} \\ &= (\cos(t) - i \sin(t)) \begin{bmatrix} -i \\ 1 \end{bmatrix} = \begin{bmatrix} -i \cos(t) - \sin(t) \\ \cos(t) - i \sin(t) \end{bmatrix} \\ &= \begin{bmatrix} -\sin(t) \\ \cos(t) \end{bmatrix} + i \begin{bmatrix} -\cos(t) \\ -\sin(t) \end{bmatrix} \\ &= \begin{bmatrix} -\sin(t) \\ \cos(t) \end{bmatrix} - i \begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} \end{aligned} \]

Complex conjugate pair again.

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We want solutions to be real-valued (no \( i \)).

Look at the real and imaginary parts:

\[ \vec{u} = \begin{bmatrix} -\sin(t) \\ \cos(t) \end{bmatrix} \quad \vec{v} = \begin{bmatrix} \cos(t) \\ \sin(t) \end{bmatrix} \]

Notice they are solutions to \( \vec{x}' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \vec{x} \)

\[ \vec{u}' = \begin{bmatrix} -\cos(t) \\ -\sin(t) \end{bmatrix} \quad \vec{u}' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} -\sin(t) \\ \cos(t) \end{bmatrix} = \begin{bmatrix} -\cos(t) \\ -\sin(t) \end{bmatrix} \]

Notice they are linearly independent:

\[ W = \begin{vmatrix} -\sin(t) & -\cos(t) \\ \cos(t) & -\sin(t) \end{vmatrix} = 1 \neq 0 \]

So, the real and imaginary parts of either solution form the fundamental solutions.

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General Solution and Phase Portraits

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

Figure: A phase portrait in the x1-x2 plane showing counter-clockwise circular orbits centered at the origin.

The origin is a center.

Another Example

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

Eigenvalues and eigenvectors:

\[ \lambda = 1 \pm i, \quad 1 - i \] \[ \vec{v} = \begin{bmatrix} 1 \\ i \end{bmatrix}, \begin{bmatrix} 1 \\ -i \end{bmatrix} \]

Form one solution:

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

Fundamental solutions

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

Figure: A phase portrait in the x1-x2 plane showing a trajectory spiraling outward from the origin.

\( e^t \) makes the ovals grow as \( t \) increases (spirals).

Origin: spiral source

Effect of the Real Part

\( e^t \): comes from the real part of eigenvalues.

Positive: spiral source (bigger spirals as \( t \) increases)
Negative: spiral sink

Figure: A phase portrait in the x1-x2 plane showing a trajectory spiraling inward toward the origin.

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

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7.8 Repeated Eigenvalues

\[ \vec{x}' = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{x} \quad \lambda = 1, 1 \]

algebraic multiplicity of two

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

\[ \left[ \begin{array}{cc|c} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \quad \vec{v} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \text{ both are free} \]

\[ \vec{v} = \begin{bmatrix} r \\ s \end{bmatrix} = r \begin{bmatrix} 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

Two linearly independent vectors

geometric multiplicity is two

if algebraic multiplicity = geometric multiplicity

the matrix is complete

solution is formed normally:

\[ \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

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star node

let's look at

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

\( \lambda = 1, 1 \) alg. mult. is two

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

\[ \left[ \begin{array}{cc|c} 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \quad \vec{v} = \begin{bmatrix} r \\ 0 \end{bmatrix} = r \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

one vector only

geo. mult. is one

if geo. mult < alg. mult

the matrix defective

\[ \vec{x}^{(1)} = e^{\lambda t} \vec{v} = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \vec{x}^{(2)} = ? \]