Question

Find \(\mathcal{L}\\{f(t)\\}\) for each of the functions \(f\) defined. $$ f(t)=\left\\{\begin{array}{ll} 0, & 02 \end{array}\right. $$

Short answer

The Laplace Transform of the given function is: \(\mathcal{L}{\\{f(t)\\}} = \frac{e^{-2s}}{s+1}\)

Step-by-step solution

Identify the two parts of the function

We have two parts for the given function: 1. \(0\) when \(02\)

Find Laplace Transform of the first part

For the first part, since the function is zero for the given interval, its Laplace Transform will also be zero, i.e., \(\mathcal{L}{\\{0\\}} = 0\)

Find Laplace Transform of the second part

To find the Laplace Transform of the second part \(e^{-t}\) when \(t>2\), we will multiply it by a unit step function, \(u(t-2)\), which equals 1 when \(t>2\) and 0 when \(t<2\). So, the second part of the function can be written as \(e^{-(t-2)}u(t-2)\), and we can now find its Laplace Transform: \(\mathcal{L}{\\{e^{-(t-2)}u(t-2)\\}} = e^{-2s}(\mathcal{L}{\\{e^{-t}\\}}) = e^{-2s}(\frac{1}{s+1})\) Here, we have used the property of Laplace Transform: \(\mathcal{L}{\\{e^{-at}f(t)\\}} = F(s+a) \), where F(s) is the Laplace Transform of f(t).

Sum the Laplace Transforms of the two parts

The overall Laplace Transform of the given function can be obtained by summing the Laplace Transforms of both parts: \(\mathcal{L}{\\{f(t)\\}} = \mathcal{L}{\\{0\\}} + \mathcal{L}{\\{e^{-(t-2)}u(t-2)\\}} = 0 + e^{-2s}(\frac{1}{s+1}) = \frac{e^{-2s}}{s+1}\) Thus, the Laplace Transform of the given function is: \[\mathcal{L}{\\{f(t)\\}} = \frac{e^{-2s}}{s+1}\]

Key concepts

Piecewise Functions

Piecewise functions are functions that have different definitions for different intervals of the domain. They're like a set of instructions telling you which rule to apply at each point on the x-axis.
In our exercise, the function is defined in two parts. This makes it a classic example of a piecewise function:

• From 0 to 2 (not including 2), the function value is 0.
• For values greater than 2, it is described by an exponential function, specifically \( e^{-t} \).

To solve such functions, you must address each part separately. Identify the intervals and apply the respective rule for that range.
This concept is essential in disciplines like physics and engineering, where different conditions can influence a system's behavior at different times.

Unit Step Function

The unit step function, often represented as \( u(t) \), is a special piecewise function. It is 0 for all negative values and 1 for non-negative values. It effectively "turns on" a function at a certain point, which is why it's so useful in defining piecewise functions over time intervals.
In our exercise, we use the unit step function \( u(t-2) \) to modify the function \( e^{-t} \) starting at \( t = 2 \).

• This means that before \( t = 2 \), the function is 0 as \( u(t-2) \) is 0.
• After \( t = 2 \), the step function shifts to 1, enabling the function to "turn on." Thus, the integral part of the function becomes \( e^{-(t-2)} \).

This technique, combining \( e^{-t} \) with \( u(t-2) \), is powerful for Laplace transforms when modeling sudden changes or events happening at a specific time.

Transform Properties

Laplace Transform has several properties that make solving differential equations more manageable. In this exercise, we leverage the property for shifting in the time domain, known as the second shifting theorem.
When we have a function \( f(t-a)u(t-a) \), where \( u(t-a) \) is a unit step function, and we need its Laplace Transform:

• We use the formula: \( \mathcal{L} \{ f(t-a)u(t-a) \} = e^{-as}F(s) \).
• Here, \( F(s) \) is the Laplace Transform of \( f(t) \).
• This property allows you to shift the function by \( a \) time units and then find the corresponding transform.

Applying this to our exercise, we determine \( \mathcal{L} \{ e^{-(t-2)} \} \) as \( e^{-2s} \frac{1}{s+1} \), using the standard form \( \mathcal{L} \{ e^{-t} \} = \frac{1}{s+1} \).
Understanding and applying these properties significantly ease the process of working through complex problems.

Differential Equations

Differential equations involve functions and their derivatives, and they arise frequently in various fields such as physics, engineering, and finance. To solve them, the Laplace Transform is a robust tool.
By transforming functions from the time domain into the frequency domain, we turn complex differential equations into simpler algebraic equations.

• Laplace Transforms convert derivative functions into polynomial forms, making calculations more straightforward.
• This allows us to focus on algebraic manipulation rather than differential calculus.
• Finally, after solving, we can apply the inverse Laplace Transform to travel back to the time domain and find the original solution.

Piecewise functions in differential equations, especially those involving the unit step function, allow us to model and solve systems with varying conditions across time intervals efficiently.