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7.8 (continued)

last time:

\[ \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} \]
\[ \lambda = 1, 1 \]
\[ \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{missing one eigenvector} \]
\[ \vec{x}^{(1)} = e^{\lambda t} \vec{v} = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \vec{x}^{(2)} = ? \]

revisit scalar case:

\[ y'' + 2y' + y = 0 \]
\[ r^2 + 2r + 1 = 0 \quad r = -1, -1 \]
\[ y = c_1 e^{-t} + c_2 t e^{-t} \]

that doesn't work for \( \vec{x}' = A \vec{x} \)

from above, let's try \( \vec{x}^{(2)} = t e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) does this satisfy \( \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} \)?

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\[ \vec{x}' = (t e^t + e^t) \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} t e^t + e^t \\ 0 \end{bmatrix} \]
\[ \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} \]
\[ \begin{bmatrix} t e^t + e^t \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} t e^t \\ 0 \end{bmatrix} = \begin{bmatrix} t e^t \\ 0 \end{bmatrix} \quad \text{NO.} \]

so, \( \vec{x}^{(2)} = t \vec{x}^{(1)} \) does not work

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

Suppose \( \vec{x}^{(1)} = e^{\lambda t} \vec{v} \) seek \( \vec{x}^{(2)} \)

try \( \vec{x}^{(2)} = t e^{\lambda t} \vec{v} \) and we'll get a clue on what's missing

\( \vec{x}' = (t \lambda e^{\lambda t} + e^{\lambda t}) \vec{v} \)

sub into \( \vec{x}' = A \vec{x} \)

\[ (t \lambda e^{\lambda t} + e^{\lambda t}) \vec{v} = A t e^{\lambda t} \vec{v} \]
\[ \lambda t e^{\lambda t} \vec{v} + e^{\lambda t} \vec{v} = t e^{\lambda t} A \vec{v} \]
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Solving Systems with Repeated Eigenvalues

\[ \begin{aligned} te^{\lambda t} \text{ terms: } & \lambda \vec{v} = A \vec{v} & \text{def. of eigenvalue and eigenvector } \vec{v} \neq 0 \\ e^{\lambda t} \text{ terms: } & \vec{v} = \vec{0} & \text{Contradiction!} \end{aligned} \]

clue on how to fix → have something on the right, so \( \vec{v} \neq 0 \)

\[ \lambda t e^{\lambda t} \vec{v} + e^{\lambda t} \vec{v} = A t e^{\lambda t} \vec{v} \]

New Solution Attempt

new solution 2: \( \vec{x}^{(2)} = t e^{\lambda t} \vec{v} + \vec{u} e^{\lambda t} \)
Where \( t e^{\lambda t} \vec{v} \) is related to \( \vec{x}^{(1)} \)

plug into \( \vec{x}' = A \vec{x} \)

repeat process above

\[ t \lambda e^{\lambda t} \vec{v} + e^{\lambda t} \vec{v} + \lambda e^{\lambda t} \vec{u} = A t e^{\lambda t} \vec{v} + A e^{\lambda t} \vec{u} \]
\[ \begin{aligned} te^{\lambda t} \text{ terms: } & \lambda \vec{v} = A \vec{v} & \text{nothing new } & (A - \lambda I) \vec{v} = \vec{0} \\ e^{\lambda t} \text{ terms: } & \vec{v} + \lambda \vec{u} = A \vec{u} & \rightarrow & \boxed{(A - \lambda I) \vec{u} = \vec{v}} \end{aligned} \]
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Generalized Eigenvectors

\( \vec{u} \) is called a generalized eigenvector

  • it gets mapped to a true eigenvector by \( (A - \lambda I) \)
  • (true eigenvectors get mapped to \( \vec{0} \))

Example Application

back to \( \vec{x}' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \vec{x} \)

\[ \begin{aligned} & \lambda = 1, 1 \quad \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ & \vec{x}^{(1)} = e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ & \vec{x}^{(2)} = t e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^t \vec{u} \end{aligned} \]

such that \( (A - \lambda I) \vec{u} = \vec{v} \)

\[ \left[ \begin{array}{cc|c} 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right] \quad \vec{u} = \begin{bmatrix} r \\ 1 \end{bmatrix} \quad \text{choose } r = 0 \\ \vec{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]
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\[ \vec{x}^{(2)} = t e^t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^t \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

General Solution:

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

Where \( e^{\lambda t} \vec{v} \) is the first term and \( t e^{\lambda t} \vec{v} + \vec{u} \) is the second term, where \( (A - \lambda I) \vec{u} = \vec{v} \).

\[ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} c_1 e^t + c_2 t e^t \\ c_2 e^t \end{bmatrix} \]

As \( t \to \infty \), \( x_1 \to \infty \), \( x_2 \to \infty \).

As \( t \to -\infty \), \( \vec{x} = \vec{0} \) but how?

\[ \lim_{t \to -\infty} \frac{x_2}{x_1} = \lim_{t \to -\infty} \frac{c_2 e^t}{c_1 e^t + c_2 t e^t} = \lim_{t \to -\infty} \frac{c_2}{c_1 + c_2 t} = 0 \text{ line of 0 slope} \]

True eigenvector \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) is line of zero slope!

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Solutions go to \( \pm \infty \) (depend on \( c_1, c_2 \)) as \( t \to \infty \) w/o following any particular line

but they go to origin along the true eigenvector (so generalized eigenvector is "invisible")

Phase portrait showing trajectories curving toward the origin along the x1 axis, which is the true eigenvector.
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7.7 Fundamental Matrix

\[ \vec{x}' = \begin{bmatrix} 3 & -2 \\ 2 & -2 \end{bmatrix} \vec{x} \quad \quad \begin{aligned} \vec{x}^{(1)} &= e^{2t} \begin{bmatrix} 2 \\ 1 \end{bmatrix} \\ \vec{x}^{(2)} &= e^{-t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{aligned} \]

put solutions into a matrix as columns

\[ \begin{aligned} \vec{x} &= c_1 \vec{x}^{(1)} + c_2 \vec{x}^{(2)} \\ &= \underbrace{\begin{bmatrix} 2e^{2t} & e^{-t} \\ e^{2t} & 2e^{-t} \end{bmatrix}}_{\Psi(t) \text{ : fundamental matrix}} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \end{aligned} \]
\[ \vec{x} = \Psi(t) \vec{c} \]
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\( \vec{c} \) comes from initial conditions

\[ \vec{x}(t_0) = \begin{bmatrix} x_1(t_0) \\ x_2(t_0) \end{bmatrix} \]
\[ \begin{aligned} \vec{x} &= \Psi(t) \vec{c} \\ \vec{x}(t_0) &= \Psi(t_0) \vec{c} \quad \text{solve for } \vec{c} \\ \vec{c} &= \Psi^{-1}(t_0) \vec{x}(t_0) \quad \Psi \text{ is invertible because solutions are linearly indp } W(\vec{x}^{(1)}, \vec{x}^{(2)}) \neq 0 \end{aligned} \]
\[ \vec{x} = \Psi(t) \Psi^{-1}(t_0) \vec{x}(t_0) \quad \rightarrow \quad \vec{x} = \Phi(t) \vec{x}(t_0) \]
\[ \Phi(t) = \Psi(t) \Psi^{-1}(t_0) \]

also a fundamental matrix

"state transition matrix"

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Let \( t_0 = 0 \)

\[ \vec{x}(t) = \Phi(t) \vec{x}(0) \]
\[ \vec{x}(t) = \Phi(t) \vec{x}_0 \]

\( \vec{x}' = A \vec{x} \)

Comparison to Scalar Case

Compare to \( y' = ay \), \( y(0) = y_0 \)

\[ \begin{aligned} y(t) &= Ce^{at} \\ y_0 &= C \\ y(t) &= y_0 e^{at} \end{aligned} \]
\[ y(t) = e^{at} y_0 \]

Note: These forms are considered the same.

So,

\( e^{At} = \Phi(t) \)