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Exam 1 Review

exam format: 8 questions (12 pts each, 4 bonus)

"hybrid multiple-choice"

25% on answer
75% on work

( 4" \( \times \) 6" ) note card

handwritten
no sharing
must be 2D

Linear
nonlinear
Laplace

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

\( \lambda = 4, 4 \)

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

\[ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \quad \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

missing one eigenvector

generalized eigenvector: \( (A - \lambda I) \vec{u} = \vec{v} \)

\[ \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \quad \vec{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

solution: \[ \vec{x} = c_1 e^{4t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{4t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right) \]

phase diagram

Figure: A phase portrait for a 2D system of differential equations on a Cartesian plane with axes x1 and x2. The origin is an improper nodal source. Trajectories curve away from the origin. Along the x1-axis, arrows point outward from the origin in both directions. In the upper half-plane, trajectories curve upwards and away from the x2-axis. In the lower half-plane, trajectories curve downwards and away from the x2-axis. All paths originate from the origin and move outwards.

origin is a nodal source (improper)

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\[ \vec{x}' = \begin{bmatrix} 1 & -2 \\ 5 & -1 \end{bmatrix} \vec{x} \quad \lambda = 3i, -3i \quad \vec{v} = \begin{bmatrix} 2 \\ 1-3i \end{bmatrix}, \begin{bmatrix} 2 \\ 1+3i \end{bmatrix} \]

sketch phase diagram

\( \lambda \)'s are purely imaginary: ovals

orientation: \[ \begin{bmatrix} 2 \\ 1-3i \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + i \begin{bmatrix} 0 \\ -3 \end{bmatrix} \]

Figure: Phase diagram on a Cartesian plane showing a blue elliptical trajectory centered at the origin. A green vector points from the origin to the point (2, 1) and is labeled with the vector [2; 1]. A red vector points from the origin to the point (0, -3) and is labeled with the vector [0; -3]. The ellipse is contained within a dashed black rectangular box.

direction: \[ \vec{x}' = \begin{bmatrix} 1 & -2 \\ 5 & -1 \end{bmatrix} \vec{x} \quad \text{pick } \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \vec{x}' = \begin{bmatrix} 1 \\ 5 \end{bmatrix} \]

right, up

counterclockwise

\( \therefore \) origin is a center

stable

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nonlinear systems

\( x' = x(1-y) \)
\( y' = y(x-3) \)

predator-prey

\( x' = x - xy \)
\( y' = -3y + xy \)

Critical pts:

\( x' = 0 \) AND \( y' = 0 \)

\( x' = 0 \rightarrow x = 0, y = 1 \)

\( y' = 0 \rightarrow y = 0, x = 3 \)

\( (0, 0), (3, 1) \)

Jacobian matrix

\( x' = f(x, y) \)
\( y' = g(x, y) \)

\[ J(x, y) = \begin{bmatrix} f_x & f_y \\ g_x & g_y \end{bmatrix} \]

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here, \( J = \begin{bmatrix} 1-y & -x \\ y & x-3 \end{bmatrix} \)

\( J(0,0) = \begin{bmatrix} 1 & 0 \\ 0 & -3 \end{bmatrix} \quad \lambda = 1, -3 \)

(0,0) is a saddle pt
unstable
solutions that start near (0,0) go away from (0,0)

\( J(3,1) = \begin{bmatrix} 0 & -3 \\ 1 & 0 \end{bmatrix} \quad \lambda \)'s are purely imaginary

(3,1) is a center
solutions that start near (3,1) orbit (3,1)

Figure: A phase plane diagram for a cooperation model in the first quadrant. The x-axis and y-axis are shown. There are fixed points at (0,1) on the y-axis and (2,0) on the x-axis. A red dot represents a stable coexistence equilibrium at (4,3). Arrows indicate that trajectories from (0,1), (2,0), and other points in the quadrant converge toward (4,3).

has higher pops than if only one species is around
both helped by the other
→ cooperation

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unit step and impulse

\[ f(t) = \begin{cases} 0 & 0 < t < \pi \\ \sin(t) & t \ge \pi \end{cases} \]

\( F(s) = ? \)

Figure: A graph of the function f(t) on a coordinate system. The horizontal axis is t. For the interval from 0 to pi, the function is zero, indicated by a blue line on the axis. Starting at t = pi, a green sine wave begins its oscillation.

\( = u_{\pi}(t) \cdot \sin(t) \)

shift LEFT by \( \pi \): \( t \to t + \pi \)

\[ F(s) = e^{-\pi s} \mathcal{L} \{ \sin(t + \pi) \} \]

\( \sin(t) \) has period of \( 2\pi \)

\( \sin(t + \pi) \) shifted by half period \( = -\sin(t) \)

Figure: A graph of a standard sine wave over one period from 0 to 2 pi. The first half-cycle from 0 to pi is blue, and the second half-cycle from pi to 2 pi is highlighted in red, illustrating that it is the negative of the first half.

\[ F(s) = e^{-\pi s} \mathcal{L} \{ -\sin(t) \} \]

\[ = e^{-\pi s} \cdot \frac{-1}{s^2 + 1} \]

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convolution:

\[ \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} f(t-\tau) g(\tau) d\tau = f * g \]

\[ \mathcal{L}\{f * g\} = \mathcal{L}\{f\} \mathcal{L}\{g\} = FG \]

\[ \int_{0}^{t} \underbrace{(t-\tau)^2}_{f(t-\tau)} \underbrace{\delta(\tau-3)}_{g(\tau)} d\tau = ? \]

in S-domain: \(f(t) = t^2\), \(g(t) = \delta(t-3)\)

\(f(t) = t^2 \rightarrow \frac{2}{s^3}\)
\(g(t) = \delta(t-3) \rightarrow e^{-3s}\)

\[ e^{-3s} \left( \frac{2}{s^3} \right) \]

shift RIGHT by 3: \(t \rightarrow t-3\)

back to t: \(u_3(t) \cdot (t-3)^2\)

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\[ y'' + 4y = \delta(t - \pi) \quad y(0) = y'(0) = 0 \]

\[ s^2 Y + 4Y = e^{-\pi s} \]

\[ (s^2 + 4)Y = e^{-\pi s} \]

\[ Y = e^{-\pi s} \cdot \frac{1}{s^2 + 4} \]

\[ y = u_{\pi}(t) \cdot \frac{1}{2} \sin(2(t - \pi)) \]