Problem 1 - Advanced Engineering Mathematics

Question

Use the definition of the Laplace transform to find \(\mathscr{L}\\{f(t)\\}\). $$ f(t)=\left\\{\begin{array}{lr} t, & 0 \leq t<1 \\ 2-t, & t \geq 1 \end{array}\right. $$

Step-by-Step Solution

Answer

The Laplace transform of \( f(t) \) is \( \frac{1 - e^{-s}}{s^2} + \frac{e^{-s}}{s}(2s - 1) \).

Step 1: Understanding the Laplace Transform Definition

The Laplace transform of a function \( f(t) \) is defined as \( \mathscr{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} \, dt \). This is an integral transform where \( s \) is a complex number.

Step 2: Split the Function Based on the Domain

The given piecewise function \( f(t) \) has two parts: \( f(t) = t \) for \( 0 \leq t < 1 \) and \( f(t) = 2-t \) for \( t \geq 1 \). We need to evaluate the Laplace transform for each part separately and then add them together.

Step 3: Calculate Laplace Transform for the First Interval

For \( 0 \leq t < 1 \), consider \( f(t) = t \). The Laplace transform is \( \int_0^1 t e^{-st} \, dt \). Use integration by parts to solve this integral. Let \( u = t \) and \( dv = e^{-st} \, dt \). Thus, \( du = dt \) and \( v = \frac{-1}{s}e^{-st} \).

Step 4: Compute Result for First Integral using Integration by Parts

Applying integration by parts: \( \int t e^{-st} \, dt = \left[-\frac{t}{s} e^{-st} \right]_0^1 + \frac{1}{s} \int_0^1 e^{-st} \, dt \). The first term evaluates to \( -\frac{1}{s^2}(e^{-s} - 1) \). The remaining integral \( \int_0^1 e^{-st} \, dt \) evaluates to \( \frac{1}{s}(1 - e^{-s}) \). Combine these results to get the complete solution for this part.

Step 5: Calculate Laplace Transform for the Second Interval

For \( t \geq 1 \), consider \( f(t) = 2-t \). The Laplace transform is \( \int_1^{\infty} (2-t) e^{-st} \, dt \). Split this into two integrals: \( 2\int_1^{\infty} e^{-st} \, dt - \int_1^{\infty} t e^{-st} \, dt \).

Step 6: Evaluate Each Part of the Second Interval

The first integral is \( 2\left[\frac{-1}{s} e^{-st} \right]_1^{\infty} = \frac{2e^{-s}}{s} \). For the second part, using integration by parts with \( u = t \), \( dv = e^{-st} \), \( du = dt \), and \( v = \frac{-1}{s}e^{-st} \), compute \( \left[-\frac{t}{s} e^{-st} \right]_1^{\infty} - \frac{1}{s} \int_1^{\infty} e^{-st} \, dt = \frac{e^{-s}}{s^2} + \frac{1}{s^2} \).

Step 7: Combine Results for Overall Laplace Transform

Add the solutions for the two intervals: \( \mathscr{L}\{f(t)\} = -\frac{1}{s^2}(e^{-s} - 1) + \frac{1}{s^2}(1 - e^{-s}) \) for \( 0 \leq t < 1 \) and \( \frac{2e^{-s}}{s} - \frac{e^{-s}}{s^2} - \frac{1}{s^2} \) for \( t \geq 1 \). Simplify to get the final Laplace transform of \( f(t) \).

Key Concepts

Piecewise Function

A piecewise function is a function that is defined by multiple sub-functions, each of which applies to a certain interval of the main function's domain. This type of function can behave differently depending on the input value. In the context of the given exercise, the function is defined as:

• \( f(t) = t \) when \( 0 \leq t < 1 \)
• \( f(t) = 2 - t \) when \( t \geq 1 \)

In these scenarios, we must consider each piece of the domain separately when applying operations like integration. This separation allows us to approach and solve each interval according to its specific behavior. Knowing when and how to divide a piecewise function into simpler parts can simplify complex calculations and help in finding solutions to integral transforms, such as the Laplace transform.

Integration by Parts

Integration by parts is a powerful tool in calculus, especially useful when dealing with the integrals of products of functions. The formula comes from the product rule for differentiation and is written as: \[ \int u\, dv = uv - \int v\, du \]Where \( u \) and \( dv \) are differentiable functions of a single variable. In this exercise, integration by parts is used in several steps to compute the Laplace transform of the function. When applying integration by parts:

• First, identify parts of your function as \( u \) and \( dv \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Substitute these into the formula to simplify the integral.

This method helps in breaking down a complicated integral into simpler parts that are easier to solve. For instance, in the solution, functions were chosen precisely so that the integration yielded results that could be evaluated or further integrated with ease.

Integral Transform

An integral transform is a mathematical operation that converts a function from its original domain (often time or space) into another domain. This is done using an integral formula. In the context of engineering and physics, the Laplace transform is among the most prominent integral transforms.The Laplace transform is useful for analyzing systems and solving differential equations. It transforms a time-based function, \( f(t) \), into a complex frequency domain function, \( \mathscr{L}\{f(t)\} \). The definition of the Laplace transform is: \[ \mathscr{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} \, dt \]Where \( s \) is a complex number. This transformation provides benefits such as:

• Simplifying the process of solving linear differential equations.
• Allowing operations like differentiation and integration to become algebraic manipulation in the transformed domain.
• Helping in control system analysis by understanding system properties like stability.

Learning how to compute and apply the Laplace transform opens a wide range of techniques for handling complex real-world problems in various fields.