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K. Fourth-Order Differential Equation

\[ y^{(4)} = -P \quad P > 0 \quad L > 0 \] \[ y(0) = y(L) = 0 \] \[ y'(0) = y'(L) = 0 \]

Figure: Diagram of a beam between x=0 and x=L showing downward deflection y under a load.

a). Solution for Specific Parameters

Given \( L = 4 \) and \( P = 24 \):

\[ y^{(4)} = 0 \rightarrow r = 0, 0, 0, 0 \] \[ y = c_1 + c_2 t + c_3 t^2 + c_4 t^3 + Y \quad \text{where } Y = At^4 \] \[ Y' = 4At^3 \quad Y'' = 12At^2 \quad Y''' = 24At \quad Y^{(4)} = 24A \]

Substituting back into the differential equation:

\[ 24A = -P \implies A = -\frac{P}{24} = -\frac{24}{24} = -1 \]

The general solution becomes:

\[ y = c_1 + c_2 t + c_3 t^2 + c_4 t^3 - t^4 \]

Applying boundary conditions:

\( 0 = c_1 + 0 \rightarrow c_1 = 0 \)
\( 0 = c_1 + 4c_2 + 16c_3 + 64c_4 - 256 \dots \)

b). Maximum Displacement

To find the maximum displacement:

max displacement \( \rightarrow \) max \( y \rightarrow y' = 0 \)

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6.1 The Laplace Transform

The Laplace transform of \( f(t) \) is defined as:

\[ \mathcal{L} \{ f(t) \} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt \]

Named after Pierre-Simon de Laplace (1749–1827)
Application to DEs is due to Oliver Heaviside (1850–1925)

Conditions for Existence

\( f(t) \) must be at least piecewise continuous:

→ finite number of discontinuities
→ the limit of \( f(t) \) at each end of an subinterval must be finite.

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Piecewise Continuity

Example of a Piecewise Continuous Function

A function is piecewise continuous if it has a finite number of jump discontinuities on any finite interval and is continuous elsewhere.

piecewise continuous

Example of a Non-Piecewise Continuous Function

NOT piecewise continuous because it has an infinite limit at at least one end of a subinterval.

In this case, the function approaches infinity as it nears the vertical asymptotes at \( t = -\pi/2 \) and \( t = \pi/2 \).

Figure: Graph of a function with vertical asymptotes at -pi/2 and pi/2, showing it is not piecewise continuous.

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The Laplace Transform

Definition

\[ \mathcal{L} \{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt \]

Example: Constant Function

Let \( f(t) = 1 \). We calculate its Laplace transform:

\[ \begin{aligned} F(s) &= \int_{0}^{\infty} e^{-st} dt \\ &= \lim_{A \to \infty} \int_{0}^{A} e^{-st} dt \\ &= \lim_{A \to \infty} \left. -\frac{1}{s} e^{-st} \right|_{0}^{A} \\ &= \lim_{A \to \infty} \left( -\frac{1}{s} e^{-sA} + \frac{1}{s} \right) \end{aligned} \]

Convergence Condition:

The integral converges if \( s > 0 \). (Otherwise, the first term blows up).

\[ \mathcal{L} \{1\} = \frac{1}{s} , \quad s > 0 \]

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Example

LIATE

Let \( f(t) = t \)

\[ \begin{aligned} \mathcal{L} \{ t \} &= \int_{0}^{\infty} e^{-st} \cdot t \, dt \\ &= \lim_{A \to \infty} \int_{0}^{A} t e^{-st} \, dt \end{aligned} \]

\( u = t \)
\( du = dt \)

\( dv = e^{-st} dt \)
\( v = -\frac{1}{s} e^{-st} \)

\[ \begin{aligned} &= \lim_{A \to \infty} \left( -\frac{t}{s} e^{-st} \Big|_0^A + \frac{1}{s} \int_{0}^{A} e^{-st} \, dt \right) \\ &= \lim_{A \to \infty} \left( -\frac{A}{s} e^{-sA} - \frac{1}{s^2} e^{-st} \Big|_0^A \right) \\ &= \lim_{A \to \infty} \left( -\frac{A}{s} e^{-sA} - \frac{1}{s^2} e^{-sA} + \frac{1}{s^2} \right) \quad \text{for } s > 0 \end{aligned} \]

\( = \frac{1}{s^2}, \quad s > 0 \)

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Example

\[ \begin{aligned} &\mathcal{L} \{ 1 + t \} \\ &= \int_{0}^{\infty} (1 + t) e^{-st} \, dt \\ &= \int_{0}^{\infty} 1 \cdot e^{-st} \, dt + \int_{0}^{\infty} t \cdot e^{-st} \, dt \\ &= \mathcal{L} \{ 1 \} + \mathcal{L} \{ t \} \end{aligned} \]

so, \( \mathcal{L} \{ f(t) \pm g(t) \} = \mathcal{L} \{ f(t) \} \pm \mathcal{L} \{ g(t) \} \)

also, \( \mathcal{L} \{ c \cdot f(t) \} = c \cdot \mathcal{L} \{ f(t) \} \)

\( \uparrow \)
constant

LT is linear

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Example: Laplace Transform of \(\sin(bt)\)

Let \(b\) be a constant.

Using Euler's formula, we can express the sine function as:

\[\sin(bt) = \frac{e^{ibt} - e^{-ibt}}{2i} = \frac{1}{2i} (e^{ibt} - e^{-ibt})\]

The Laplace transform is then:

\[\mathcal{L} \left\{ \frac{1}{2i} (e^{ibt} - e^{-ibt}) \right\} = \frac{1}{2i} (\mathcal{L} \{e^{ibt}\} - \mathcal{L} \{e^{-ibt}\})\]

\[= \frac{1}{2i} \int_{0}^{\infty} e^{-st} (e^{ibt} - e^{-ibt}) dt = \frac{1}{2i} \lim_{A \to \infty} \int_{0}^{A} e^{-st} (e^{ibt} - e^{-ibt}) dt\]

\[= \frac{1}{2i} \lim_{A \to \infty} \int_{0}^{A} (e^{(ib-s)t} - e^{-(ib+s)t}) dt\]

\[= \frac{1}{2i} \lim_{A \to \infty} \left[ \frac{1}{ib-s} e^{(ib-s)t} \bigg|_0^A + \frac{1}{ib+s} e^{-(ib+s)t} \bigg|_0^A \right]\]

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\[= \frac{1}{2i} \lim_{A \to \infty} \left[ \frac{1}{ib-s} (e^{(ib-s)A} - 1) + \frac{1}{ib+s} (e^{-(ib+s)A} - 1) \right]\]

Recall: \(e^{ibt} = \cos(bt) + i \sin(bt)\). Its magnitude is bounded.

As \(A \to \infty\), the terms \(e^{(ib-s)A} = e^{ibA} e^{-sA}\) and \(e^{-(ib+s)A} = e^{-ibA} e^{-sA}\) both go to 0 if \(s > 0\).

\[= \frac{1}{2i} \left( \frac{1}{s-ib} - \frac{1}{s+ib} \right) = \frac{1}{2i} \frac{2ib}{s^2+b^2} = \frac{b}{s^2+b^2}\]

\(\frac{b}{s^2+b^2}\)

\(s > 0\)

2nd way:

Evaluate the integral directly:

\[\int_{0}^{\infty} e^{-st} \sin(bt) dt\]

Integrate by parts twice.

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Computer Project 2. RLC Circuits

7/18

Goal: Investigate the charge on a capacitor in an RLC circuit with varying voltage.

Tools needed: ode45, plot

Description: If \( Q(t) \) = charge on a capacitor at time \( t \) in an RLC circuit (with \( R, L \) and \( C \) being the resistance, inductance and capacitance, respectively) and \( E(t) \) = applied voltage, then Kirchhoff's Laws give the following \( 2^{\text{nd}} \) order differential equation for \( Q(t) \):

\[ L Q''(t) + R Q'(t) + \frac{1}{C} Q(t) = E(t) \tag{*} \]

Figure: Schematic of an RLC circuit with a voltage source E(t), resistor R, inductor L, and capacitor C in a single loop.

Questions: Assume \( L = 1, C = 1/5, R = 4 \) and \( E(t) = 10 \cos \omega t \).

Use ode45 (and plot routines) to plot the solution of (*) with \( Q(0) = 0 \) and \( Q'(0) = 0 \) over the interval \( 0 \le t \le 80 \) for \( \omega = 0, 0.5, 1, 2, 4, 8, 16 \).
Let \( A(\omega) = \text{maximum of } |Q(t)| \) over the interval \( 30 \le t \le 80 \) (this approximates the amplitude of the steady-stat solution). Experiment with various values of \( \omega \) and discuss what appears to happen to \( A(\omega) \) as \( \omega \to \infty \) and as \( \omega \to 0 \). Also, interpret your findings in terms of an equivalent spring-mass system.

Remark: There is an analogy between spring-mass system and RLC circuits given by:

Spring-mass system	RLC circuit
\( m u'' + c u' + k u = F(t) \)	\( L Q'' + R Q' + \frac{1}{C} Q = E(t) \)
\( u = \text{Displacement} \)	\( Q = \text{Charge} \)
\( u' = \text{Velocity} \)	\( Q' = I = \text{Current} \)
\( m = \text{Mass} \)	\( L = \text{Inductance} \)
\( c = \text{Damping constant} \)	\( R = \text{Resistance} \)
\( k = \text{Spring constant} \)	\( 1/C = (\text{Capacitance})^{-1} \)
\( F(t) = \text{External force} \)	\( E(t) = \text{Voltage} \)