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6.4 Diff. Eqs. with Discontinuous Forcing Functions

from last time: \( u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \ge c \end{cases} \)

\[ \mathcal{L} \{ u_c(t) f(t-c) \} = e^{-cs} \mathcal{L} \{ f(t) \} \]

function activated by \( u_c \) but shifted back to origin \( t \to t+c \)

now we can solve something like

\[ y'' + 3y' + 2y = \begin{cases} 0 & 0 \le t < \pi \\ 1 & \pi \le t < 2\pi \\ 0 & t \ge 2\pi \end{cases} \quad y(0)=0, \, y'(0)=1 \]

interpretation: mass-spring-damper \( m=1, \, \gamma=3, \, k=2 \)

Figure: Diagram of a mass-spring-damper system with mass m=1, damping gamma=3, and spring constant k=2.

\( F(t) \) external force

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w/o using Laplace transform:

first, solve \( y'' + 3y' + 2y = 0 \) with \( y(0)=0, \, y'(0)=1 \) for \( 0 \le t < \pi \)
then, solve \( y'' + 3y' + 2y = 1 \) with \( y(\pi) = \text{---}, \, y'(\pi) = \text{---} \) for \( \pi \le t < 2\pi \)
from solution of 1st part at \( t = \pi \)
then, solve \( y'' + 3y' + 2y = 0 \) with \( y(2\pi) = \text{---}, \, y'(2\pi) = \text{---} \) for \( t \ge 2\pi \)
come from \( y(2\pi) \) from 2nd part

with Laplace transform, we do all that at once

\[ F(t) = \begin{cases} 0 & 0 \le t < \pi \\ 1 & \pi \le t < 2\pi \\ 0 & t \ge 2\pi \end{cases} \]

Figure: Graph of the piecewise forcing function F(t) showing a pulse of height 1 between pi and 2pi.

in terms of \( u_c(t) \)

\[ = u_{\pi}(t) - u_{2\pi}(t) \]

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Solving Differential Equations with Laplace Transforms

\[ y'' + 3y' + 2y = u_{\pi} - u_{2\pi} \quad y(0)=0, \ y'(0)=1 \]

Laplace Transform

\[ s^2 Y - s y(0) - y'(0) + 3 [s Y - y(0)] + 2Y = e^{-\pi s} \frac{1}{s} - e^{-2\pi s} \frac{1}{s} \]

Note: \(y(0)=0\) (given), \(y'(0)=1\) (given).

\[ (s^2 + 3s + 2) Y = 1 + e^{-\pi s} \frac{1}{s} - e^{-2\pi s} \frac{1}{s} \]

Characteristic equation: \(s^2 + 3s + 2\)

\[ Y = \frac{1}{s^2 + 3s + 2} + e^{-\pi s} \frac{1}{s(s^2 + 3s + 2)} - e^{-2\pi s} \frac{1}{s(s^2 + 3s + 2)} \]

Inverse Laplace Transform

\[ \frac{1}{s^2 + 3s + 2} = \frac{1}{(s+2)(s+1)} = \frac{A}{s+2} + \frac{B}{s+1} = \dots = \frac{1}{s+1} - \frac{1}{s+2} \]

\[ \frac{1}{s(s^2 + 3s + 2)} = \frac{1}{(s)(s+2)(s+1)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+1} = \dots = \frac{1}{2} \frac{1}{s} - \frac{1}{s+1} + \frac{1}{2} \frac{1}{s+2} \]

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\[ Y = \left( \frac{1}{s+1} - \frac{1}{s+2} \right) + e^{-\pi s} \left( \frac{1}{2} \frac{1}{s} - \frac{1}{s+1} + \frac{1}{2} \frac{1}{s+2} \right) - e^{-2\pi s} \left( \frac{1}{2} \frac{1}{s} - \frac{1}{s+1} + \frac{1}{2} \frac{1}{s+2} \right) \]

Inverse Transform Mappings:

\(\frac{1}{s+1} \to e^{-t}\)
\(\frac{1}{s+2} \to e^{-2t}\)
\(e^{-\pi s} \to u_{\pi}\) with shift right by \(\pi\) units of \(t\) (\(t \to t-\pi\))
\(e^{-2\pi s} \to u_{2\pi}\) with shift \(t \to t-2\pi\)

\[ y = e^{-t} - e^{-2t} + u_{\pi} \left( \frac{1}{2} - e^{-(t-\pi)} + \frac{1}{2} e^{-2(t-\pi)} \right) - u_{2\pi} \left( \frac{1}{2} - e^{-(t-2\pi)} + \frac{1}{2} e^{-2(t-2\pi)} \right) \]

Piecewise Definition

\[ y = \begin{cases} e^{-t} - e^{-2t} & 0 \le t < \pi \\ \frac{1}{2} + e^{-t} - e^{-2t} - e^{-(t-\pi)} + \frac{1}{2} e^{-2(t-\pi)} & \pi \le t < 2\pi \\ e^{-t} - e^{-2t} - e^{-(t-\pi)} + \frac{1}{2} e^{-2(t-\pi)} + e^{-(t-2\pi)} - \frac{1}{2} e^{-2(t-2\pi)} & t \ge 2\pi \end{cases} \]

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Differential Equation Response Visualization

The following graph illustrates the response of a system, likely an underdamped oscillator, subjected to a piecewise forcing function. The blue curve represents the system response \( y(t) \), while the red vertical lines indicate the timing of the forcing function \( F(t) \).

Figure: A plot with a blue curve y(t) and red vertical lines F(t) at t=pi and t=2pi on a time axis from 0 to 10.

Graph Analysis

The graph features two vertical axes. The left axis (blue) ranges from 0 to 0.5, corresponding to the displacement \( y(t) \). The right axis (red) ranges from 0 to 1, corresponding to the forcing function \( F(t) \).

The system starts at \( y(0) = 0 \) and reaches an initial peak near \( t = 1 \).
A forcing event occurs at \( t \approx 3.14 \) (likely \( \pi \)), causing a significant increase in amplitude.
A second forcing event occurs at \( t \approx 6.28 \) (likely \( 2\pi \)), after which the system begins to decay.

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Underdamped System Analysis

Change \( \gamma \) to 2 (now underdamped):

\[ y'' + 2y' + 2y = u_{\pi} - u_{2\pi}, \quad y(0) = 0, \, y'(0) = 1 \]

Following the same steps, we get:

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

Completing the Square

\[ \frac{1}{s^2 + 2s + 2} = \frac{1}{(s+1)^2 + 1} \]

Completed the square.

This looks a lot like \( \frac{1}{s^2 + 1} \) which is \( \mathcal{L}\{\sin(t)\} \).

Applying the s-Shifting Theorem

From last time:

\[ F(s-c) = \mathcal{L}\{e^{ct} f(t)\} \]

\[ F(s) = \frac{1}{s^2 + 1} = \mathcal{L}\{\sin(t)\} \] \[ F(s+1) = \frac{1}{(s+1)^2 + 1} = \mathcal{L}\{e^{-t} \sin(t)\} \]

Note: \( c = -1 \)

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Partial Fraction Decomposition

Middle part:

\[ \frac{1}{s(s^2+2s+2)} = \frac{A}{s} + \frac{Bs+C}{s^2+2s+2} \]

Note: The numerator \( Bs+C \) is linear (one degree lower than the denominator), and the denominator \( s^2+2s+2 \) is quadratic.

\[ = \dots = \frac{1}{2} \frac{1}{s} - \frac{1}{2} \frac{s+2}{(s+1)^2+1} \]\[ = \frac{1}{2} \frac{1}{s} - \frac{1}{2} \frac{s+1}{(s+1)^2+1} - \frac{1}{2} \frac{1}{(s+1)^2+1} \]

Applying inverse Laplace transforms:

\( \frac{s+1}{(s+1)^2+1} \rightarrow e^{-t} \cos(t) \)
\( \frac{1}{(s+1)^2+1} \rightarrow e^{-t} \sin(t) \)

Inverse Laplace Transform in the s-domain

\[ Y = \frac{1}{(s+1)^2+1} + e^{-\pi s} \left( \frac{1}{2} \frac{1}{s} - \frac{1}{2} \frac{s+1}{(s+1)^2+1} - \frac{1}{2} \frac{1}{(s+1)^2+1} \right) - e^{-2\pi s} ( \dots ) \]

For the term with \( e^{-\pi s} \), shift back to \( t \): \( t \rightarrow t - \pi \). For the term with \( e^{-2\pi s} \), the expression in parentheses is the same, shift \( t \rightarrow t - 2\pi \).

Final Solution in the Time Domain

\[ y = e^{-t} \sin(t) + u_{\pi} \left( \frac{1}{2} - \frac{1}{2} e^{-(t-\pi)} \cos(t-\pi) - \frac{1}{2} e^{-(t-\pi)} \sin(t-\pi) \right) \]\[ - u_{2\pi} \left( \frac{1}{2} - \frac{1}{2} e^{-(t-2\pi)} \cos(t-2\pi) - \frac{1}{2} e^{-(t-2\pi)} \sin(t-2\pi) \right) \]

Using trigonometric identities:

\( \cos(t-\pi) = -\cos(t) \)
\( \sin(t-\pi) = -\sin(t) \)
\( \cos(t-2\pi) = \cos(t) \)
\( \sin(t-2\pi) = \sin(t) \)