Section 4.4 Step functions
Objective:
1. Definition of the step function
2. Use Laplace transformation to solve differential equations with step function
Recall that the whole theme of this chapter is using Laplace transform to solve IVP that cannot be solved using previous chapters. So far, we have been solving IVP that can be solved using previous chapters! Here, we introduce new functions that are not continuous. We then can solve IVP involve discontinue functions. One of practical functions in the life is the step functions. We start with easy one here.
Definition: Let [latex]c \gt 0[/latex]. The unit step function is defined by
\[u_{c}(t)=\begin{cases} 0, & t \lt c\\ 1, & t\geq c \end{cases}.\]
Example 1: Write [latex]g(t)=5u_{2}(t)-u_{4}(t)+3u_{7}(t)[/latex], [latex]t\geq 0[/latex] as a piecewise function and draw the graph of [latex]g(t)[/latex].
Exercise 1: Write [latex]g(t)=3u_{2}(t)+2u_{6}(t)-4u_{9}(t)[/latex], [latex]t\geq0[/latex] as a piecewise function and draw the graph of [latex]g(t)[/latex].
Example 2: Write [latex]f(t)[/latex] as a combination of step functions and draw the graph of [latex]f(t)[/latex].
\[f(t)=\begin{cases} 3, & 0\leq t \lt 2\\ -2, & 2\leq t\leq5\\ 1, & t \gt 5 \end{cases}\]
Exercise 2: Write [latex]f(t)[/latex] as a combination of step functions and draw the graph of [latex]f(t)[/latex].
\[ f(t)=\begin{cases} -1, & 0\leq t \lt 3\\ 2, & 3\leq t\leq4\\ 3, & t \gt 4 \end{cases} \]
Corollary: The Laplace transform of [latex]u_{c}(t)[/latex] is
\[\frac{e^{-cs}}{s}.\]
Example 3: Find Laplace transform of [latex]g(t)=5u_{2}(t)-u_{4}(t)+3u_{7}(t)[/latex].
Exercise 3: Find Laplace transform of[latex]g(t)=3u_{2}(t)+2u_{6}(t)-4u_{9}(t)[/latex].
There are more functions behave similar to unit step functions. Translated function is another function that is discontinue but more general. Given a function [latex]f(t)[/latex], let
\[g(t)=\begin{cases} 0, & t \lt c\\ f(t-c), & t\geq c, \end{cases}\]
then [latex]g(t)=u_{c}(t)f(t-c)[/latex].
Corollary: Let [latex]L\{f(t)\}=F(s)[/latex], then the Laplace transform of [latex]u_{c}(t)f(t-c)[/latex] is [latex]e^{-cs}F(s)[/latex].
Example 4: Find Laplace transform of
\[ g(t)=\begin{cases} 0, & t \lt 3\\ (t-3)^{5}, & t\geq3 \end{cases} .\]
Exercise 4: Find Laplace transform of
\[g(t)=\begin{cases} 0, & t \lt 5\\ (t-5)^{4}, & t\geq5 \end{cases}\]
Example 5: Find Laplace transform of
\[g(t)=\begin{cases} 0, & t \lt e\\ (t-e), & e\leq t\leq2e\\ t & t\geq2e \end{cases}.\]
Exercise 5: Find Laplace transform of
\[g(t)=\begin{cases} 0, & t \lt \pi\\ t, & \pi\leq t\leq3\pi\\ (t-3\pi) & t\geq3\pi \end{cases}.\]
We can do the translated function to the function of Laplace transform functions.
Corollary: Let [latex]L\{f(t)\}=F(s)[/latex], then the Laplace transform of [latex]e^{ct}f(t)[/latex] is [latex]F(s-c)[/latex] for [latex]s \gt a+c[/latex].
Example 6: Find Laplace transform of [latex]e^{-2t}t^{2}[/latex].
Exercise 6: Find Laplace transform of [latex]e^{3t}\text{sin}(2t)[/latex].
Example 7: Find inverse Laplace transform of
\[\frac{e^{-2s}s}{(s^{2}-2s+5)(s+1)}\]
Exercise 7: Find the inverse Laplace transform of
\[\frac{e^{-3s}s}{(s^{2}-4s+13)(s-1)}\]
Group Work
1. Use Laplace Transform to solve the IVP. [latex]y''+9y=2u_{e}(t)+u_{2e}(t)[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=0[/latex].
2. Use Laplace Transform to solve the IVP. [latex]y''+4y=\text{cos}(t)-u_{e}(t)\text{cos}(t-e)[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=0[/latex].
3. Use Laplace Transform to solve the IVP. [latex]y''+y=f(t)[/latex], [latex]y(0)=0[/latex],
[latex]y'(0)=1[/latex], where
\[f(t)=\begin{cases} 1, & 0\leq t\leq3\pi\\ 0, & 3\pi\leq t \end{cases}\]