MA 30300: Differential Equations

7.5. Piecewise Continuous Input Functions

Goal: Learn how to construct piecewise functions using unit step functions and apply them to solve differential equations.

1. Building Piecewise Functions with Unit Step Functions

Unit Step Function

Definition: The unit step function is defined by

\[ u(t - a) = \begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases} \]

Laplace Transform: The Laplace transform of the unit step function is

\[ \mathcal{L}\{u(t - a)\} = \frac{e^{-as}}{s}, \quad a \ge 0 \]

Translation on the \(t\)-Axis

Theorem (Translation on the \(t\)-Axis): Suppose \( \mathcal{L}\{f(t)\} = F(s) \) exists for \( s > c \). Then for \( a > 0 \),

\[ \mathcal{L}\{u(t-a)f(t-a)\} = e^{-as}F(s) \]
\[ \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a) \]

for \( s > c + a \).

The function \( u(t-a)f(t-a) \) represents a delayed version of \( f(t) \):

\[ u(t-a)f(t-a) = \begin{cases} 0, & t < a \\ f(t-a), & t \ge a \end{cases} \]

Representing Piecewise Functions

Idea: Piecewise functions can be rewritten using unit step functions.

Example 1.1

Problem: Find the Laplace transform of the function

\[ f(t) = \begin{cases} 0, & t < 3 \\ t^2, & t \ge 3 \end{cases} \]

Solution:

  1. Rewrite \(f(t)\) using the unit step function:
    2. \[ f(t) = u(t-3)\cdot (t^2) \]

    Since the step occurs at \(t = 3\), we shift the variable to match the Translation on the \(t\)-Axis theorem:

    \[ f(t) = u(t-3) \cdot [(t-3 + 3)^2] = u(t-3) \cdot [(t-3)^2 + 6(t-3) + 9] \]
  2. Apply the Laplace transform using linearity and the Translation on the \(t\)-Axis theorem:
    3. \[ \mathcal{L}\{u(t-a)g(t-a)\} = e^{-as} \mathcal{L}\{g(t)\} \]
    we have
    \[ \mathcal{L}\{f(t)\} = e^{-3s} \mathcal{L}\{(t)^2 + 6t + 9\} = e^{-3s} \left[ \frac{2}{s^3} + \frac{6}{s^2} + \frac{9}{s} \right] \]
  3. Final Answer:
    4. \[ \boxed{\mathcal{L}\{f(t)\} = e^{-3s} \left( \frac{2}{s^3} + \frac{6}{s^2} + \frac{9}{s} \right)} \]

Practice Problems

Question 1

Find the inverse Laplace transform:

\[ \mathcal{L}^{-1}\left\{\frac{e^{-5s}}{s+5}\right\} \]

Question 2

Find the inverse Laplace transform:

\[ \mathcal{L}^{-1}\left\{\frac{e^{-\pi s}}{s^2 + 9}\right\} \]

Question 3

Find the Laplace transform of the piecewise function:

\[ f(t) = \begin{cases} \sin 6t, & 0 \le t \le 2\pi \\ 0, & t > 2\pi \end{cases} \]

Question 4

Find the Laplace transform of the piecewise function:

\[ f(t) = \begin{cases} \sin(\pi t), & 3 \le t \le 7 \\ 0, & \text{otherwise} \end{cases} \]

Question 5

Find the Laplace transform of the piecewise function:

\[ f(t) = \begin{cases} t, & 0 \le t \le 8 \\[1mm] 72 - 8t, & 8 < t \le 9 \\[1mm] 0, & t > 9 \end{cases} \]

2. Applications to Differential Equations

Example 2.1

A mass of 1 slug (\(= 32 \) lb) is attached to a spring with constant \(k = 4\) lb/ft. The mass is initially at rest at equilibrium. Beginning at time \(t = 0\) (seconds), an external force \(f(t) = \cos 2t\) is applied to the mass, but at time \(t = 2 \pi\) this force is turned off (abruptly discontinued) and the mass is allowed to continue its motion unimpeded:

\[ f(t) = \begin{cases} \cos 2t, & 0 \le t < 2\pi \\ 0, & t \ge 2 \pi \end{cases} \]

Find the resulting position \(x(t)\) of the mass for \(t \ge 0\).

Solution:

  1. Write the forcing function using unit step functions:
    2. \[ f(t) = \cos 2t - u(t-2\pi)\cos 2(t-2\pi) \]
  2. Write the differential equation for the mass-spring system:
    3. \[ m x'' + k x = f(t), \quad x(0) = 0, \quad x'(0) = 0 \] \[ \Rightarrow x'' + 4 x = f(t) \]
  3. Take the Laplace transform of both sides:
    4. \[ s^2 X(s) - s x(0) - x'(0) + 4 X(s) = \mathcal{L}\{f(t)\} = \frac{s}{s^2 + 4} - e^{-2\pi s}\frac{s}{s^2 + 4} \] \[ (s^2 + 4)X(s) = \frac{s}{s^2 + 4} - e^{-2\pi s}\frac{s}{s^2 + 4} \]
  4. Solve for \(X(s)\):
    5. \[ X(s) = \frac{s}{(s^2 + 4)^2} \left( 1 - e^{-2\pi s} \right) \]
  5. Take the inverse Laplace transform using linearity and the Second Shifting Theorem:
    6. \[ x(t) = \mathcal{L}^{-1}\left\{\frac{s}{(s^2 + 4)^2}\right\} - \mathcal{L}^{-1}\left\{ e^{-2\pi s} \frac{s}{(s^2 + 4)^2} \right\} \]
  6. Compute the first term:
    7. \[ \mathcal{L}^{-1}\Big\{\frac{s}{(s^2 + 4)^2}\Big\} = \frac{1}{4} t \sin 2t \]
  7. Use Translation on the \(t\)-Axis for the second term:
    8. \[ \mathcal{L}^{-1}\Big\{ e^{-2\pi s} \frac{s}{(s^2 + 4)^2} \Big\} = u(t-2\pi) \cdot \frac{1}{4} (t-2\pi) \sin 2(t-2\pi) \]
  8. Combine the two terms to get \(x(t)\):
    9. \[ x(t) = \frac{1}{4} t \sin 2t - u(t-2\pi) \frac{1}{4} (t-2\pi) \sin 2(t-2\pi) \]
    or equivalently
    \[ x(t) = \begin{cases} \frac{t}{4} \sin 2t, & 0 \le t < 2\pi \\[1mm] \frac{\pi}{2} \sin 2t, & t \ge 2\pi \end{cases} \]

Example 2.2

Problem: Consider a mass-spring-dashpot system with \(m = 1\), \(k = 4\), and \(c = 0\). The system is governed by the differential equation:

\[ m x'' + c x' + k x = f(t), \quad x(0) = 0, \quad x'(0) = 0 \]

The external forcing function is defined by:

\[ f(t) = \begin{cases} \sin(t), & 0 \le t \le 2\pi \\[1mm] 0, & t > 2\pi \end{cases} \]
Find the solution \(x(t)\) by the Laplace transform method.

Solution:

  1. Express the forcing function using the unit step function:
    2. \[ f(t) = \sin(t) - u(t-2\pi)\sin(t-2\pi) \]
  2. Write the differential equation explicitly:
    3. \[ x'' + 4 x = f(t), \quad x(0) = 0, \quad x'(0) = 0 \]
  3. Take the Laplace transform:
    4. \[ (s^2 + 4) X(s) = \frac{1}{s^2 + 1} - e^{-2\pi s} \frac{1}{s^2 + 1} \]
    \[ \Rightarrow \quad X(s) = \frac{1}{(s^2 + 1)(s^2 + 4)} - e^{-2\pi s} \frac{1}{(s^2 + 1)(s^2 + 4)} \]
  4. Compute the inverse Laplace transform of the first term using partial fractions and tables:
    5. \[ \mathcal{L}^{-1} \left\{ \frac{1}{(s^2 + 1)(s^2 + 4)} \right\} = \frac{1}{3} \sin t - \frac{1}{6} \sin 2t \]
  5. Use Translation on the \(t\)-Axis for the second term:
    6. \[ \mathcal{L}^{-1} \left\{ e^{-2\pi s} \frac{1}{(s^2 + 1)(s^2 + 4)} \right\} = u(t-2\pi) \left[ \frac{1}{3} \sin(t-2\pi) - \frac{1}{6} \sin 2(t-2\pi) \right] \]
  6. Combine to get the complete solution:
    7. \[ x(t) = \frac{1}{3} \sin t - \frac{1}{6} \sin 2t - u(t-2\pi) \left[ \frac{1}{3} \sin(t-2\pi) - \frac{1}{6} \sin 2(t-2\pi) \right] \]
    \[ \Rightarrow x(t) = \left( \frac{1}{3} \sin t - \frac{1}{6} \sin 2t \right) \left( 1 - u(t-2\pi) \right) \]
    Optionally, write in piecewise form:
    \[ x(t) = \begin{cases} \frac{1}{3} \sin t - \frac{1}{6} \sin 2t, & 0 \le t \le 2\pi \\[1mm] 0, & t > 2\pi \end{cases} \]

Example 2.3

Note: You are allowed to use a calculator if needed for this problem.

Problem: The values of the elements of an RLC circuit are given. Solve the initial value problem with the given impressed voltage \(e(t)\):

\[ L \frac{di}{dt} + R i + \frac{1}{C} \int_0^t i(\tau)\, d\tau = e(t), \quad i(0) = 0 \]

Given values:

  • \(L = 1\)
  • \(R = 10\)
  • \(C = 4 \cdot 10^{-2}\)
  • \(e(t) = \begin{cases} 20t, & 0 \le t < 1 \\ 0, & t \ge 1 \end{cases} \)

Practice Example Problems:

The values of mass \(m\), spring constant \(k\), and dashpot resistance \(c\) are given for each mass–spring–dashpot system. Solve the initial value problems with the corresponding external force \(f(t)\) and initial conditions.

  1. \(m=1, \; k=4, \; c=0, \quad f(t)=1 \text{ for } 0 \le t < \pi, \; f(t)=0 \text{ for } t\ge\pi, \quad x(0)=0, \; x'(0)=0\)
  2. \(m=1, \; k=4, \; c=5, \quad f(t)=1 \text{ for } 0 \le t < 2, \; f(t)=0 \text{ for } t\ge 2, \quad x(0)=0, \; x'(0)=0\)
  3. \(m=1, \; k=9, \; c=0, \quad f(t)=\sin t \text{ for } 0 \le t \le 2\pi, \; f(t)=0 \text{ for } t>2\pi, \quad x(0)=0, \; x'(0)=0\)
  4. \(m=1, \; k=4, \; c=4, \quad f(t)=t \text{ for } 0 \le t \le 2, \; f(t)=0 \text{ for } t>2, \quad x(0)=0, \; x'(0)=0\)