Question

Find the Laplace transform by the method of Example \(8.4 .1 .\) Then express the given function \(f\) in terms of unit step functions as in Eqn. (8.4.6), and use Theorem 8.4 .1 to find \(\mathcal{L}(f)\). Where indicated by , graph \(f\). $$ f(t)=\left\\{\begin{array}{ll} t, & 0 \leq t<1 ,\\\ 1, & t \geq 1. \end{array}\right. $$

Short answer

Question: Determine the Laplace transform of the function \(f(t)\) given by: $$ f(t)=\left\\{\begin{array}{ll} t, & 0 \leq t<1 ,\\\ 1, & t \geq 1. \end{array}\right. $$ Answer: The Laplace transform of the given function \(f(t)\) is: $$ \mathcal{L}(f) = \frac{1}{s^2} (1 - e^{-s}). $$

Step-by-step solution

Find the Laplace transform using the method of Example 8.4.1

We will divide the function \(f(t)\) into two parts, \(g_1(t)\) and \(g_2(t)\). $$ g_1(t) = t, \quad 0 \le t < 1, $$ $$ g_2(t) = 1, \quad t \ge 1. $$ Now, we will find the Laplace transforms of \(g_1(t)\) and \(g_2(t)\) separately. For \(g_1(t)\), we have: $$ G_1(s) = \int_{0}^{1} g_1(t)e^{-st} dt = \int_{0}^{1} te^{-st} dt. $$ To compute this integral, we will use integration-by-parts: Let \(u = t\) and \(dv = e^{-st}dt\). Then we have \(du = dt\) and \(v = -\frac{1}{s}e^{-st}\). According to integration-by-parts formula: $$ \int u dv = uv - \int v du. $$ Applying the formula for \(G_1(s)\), we get: $$ G_1(s) = -\left. \frac{1}{s}te^{-st}\right|_{0}^{1} + \frac{1}{s} \int_{0}^{1} e^{-st} dt = -\frac{1}{s}e^{-s} + \frac{1}{s^2}(1-e^{-s}). $$ For \(g_2(t)\), we have: $$ G_2(s) = \int_{1}^{\infty} g_2(t)e^{-st} dt = \int_{1}^{\infty} e^{-st} dt = -\left . \frac{1}{s} e^{-st} \right |_{1}^{\infty} = \frac{1}{s}e^{-s}. $$ Now, using the method of Example 8.4.1, we have: $$ \mathcal{L}(f) = G_1(s) + G_2(s) = -\frac{1}{s}e^{-s} + \frac{1}{s^2}(1-e^{-s}) + \frac{1}{s}e^{-s} = \frac{1}{s^2}(1-e^{-s}). $$

Express the function \(f(t)\) in terms of unit step functions and use Theorem 8.4.1

We can rewrite the function \(f(t)\) in terms of unit step functions as follows: $$ f(t) = tg_1(t) + g_2(t) = t [u(t) - u(t - 1)] + [u(t - 1)]. $$ Now using Theorem 8.4.1, we have: $$ \mathcal{L}(f) = \mathcal{L}(tu(t)) - \mathcal{L}(t u(t - 1)) + \mathcal{L}(u(t - 1)). $$ We already found: $$ \mathcal{L}(f) = \frac{1}{s^2} (1 - e^{-s}). $$ Therefore, $$ \mathcal{L}(tu(t)) - \mathcal{L}(t u(t - 1)) + \mathcal{L}(u(t - 1)) = \frac{1}{s^2} (1 - e^{-s}). $$ So, the Laplace transform of the given function \(f(t)\) is: $$ \mathcal{L}(f) = \frac{1}{s^2} (1 - e^{-s}). $$

Key concepts

Unit Step Function

The Unit Step Function, often denoted as \( u(t) \), is a basic concept in engineering and mathematics used to switch signals on and off. It essentially represents a signal that starts at a certain time and maintains a certain constant value (usually 1). This function is particularly useful for dealing with piecewise-defined functions, like the one in the exercise.
The function \( u(t) \) is defined as:

• \( u(t) = 0 \), when \( t < 0 \)
• \( u(t) = 1 \), when \( t \geq 0 \)

For transforming functions like \( f(t) \) in our exercise, which changes its nature at \( t = 1 \), we can express different segments of the function with unit step functions. For instance, \( t[u(t) - u(t-1)] \) represents the segment \( t \) from \( t = 0 \) to \( t = 1 \), and \( u(t-1) \) starts the constant part from \( t = 1 \). Integrating these uses of the unit step function simplifies the handling of piecewise functions, especially when applying the Laplace Transform.

Integration by Parts

Integration by Parts is a powerful technique derived from the product rule for differentiation. It is particularly useful when integrating the product of two functions, one of which can be easily differentiated, and the other can be easily integrated. The formula for integration by parts is:

• \( \int u \, dv = uv - \int v \, du \)

In the given problem, integration by parts was applied to compute the Laplace Transform of \( g_1(t) = t \) over \( [0,1) \). Here, we set \( u = t \) and \( dv = e^{-st} dt \), making \( du = dt \) and \( v = -\frac{1}{s}e^{-st} \).
This turns what might be a complex integral into a simpler equation. Applying the parts formula, we calculate:

• \( \int te^{-st} dt = -\left. \frac{1}{s}te^{-st}\right|_{0}^{1} + \frac{1}{s} \int_{0}^{1} e^{-st} dt \)

This approach provides a structured method for dealing with integrals in Laplace Transforms, yielding manageable functions that contribute to our final transformation.

Piecewise Functions

Piecewise functions, like \( f(t) \) in the exercise, are defined by different expressions in different intervals of their domain. In this scenario, \( f(t) \) changes its behavior at \( t = 1 \). The ability to manipulate and transform piecewise functions is vital in various fields of study, particularly in signals and systems.
Consider how \( f(t) \) is expressed:

• \( f(t) = t \) for \( 0 \leq t < 1 \)
• \( f(t) = 1 \) for \( t \geq 1 \)

When integrating or transforming such functions, we often use unit step functions to simplify complexity. Expressing piecewise functions in terms of unit steps allows for simplification when applying operations like the Laplace Transform. This technique manages different function segments seamlessly and unifies them into a single mathematical expression.

Theorem in Differential Equations

Theorems play an essential role in simplifying and solving complex mathematical problems. For this exercise, Theorem 8.4.1 offers a method to find the Laplace Transform of functions expressed with unit steps. The theorem essentially states that the Laplace Transform \( \mathcal{L}(f) \) of a piecewise function built from unit step functions can be broken down into simpler parts. The transformation process breaks into steps and combines them according to the rule:

• \( \mathcal{L}(f) = \mathcal{L}(tu(t)) - \mathcal{L}(t u(t - 1)) + \mathcal{L}(u(t - 1)) \)

This simplifies complex transformations into a manageable calculation, as evidenced by our final result of \( \mathcal{L}(f) = \frac{1}{s^2} (1 - e^{-s}) \).
By applying this theorem, we manage the interrelationship between processes acting on different function parts within the given domain. This highlights why theorems in differential equations are essential, providing structured solutions to dynamic and sophisticated mathematical challenges.