Problem 22 - Advanced Engineering Mathematics

Question

From the Table of Laplace transforms we have $$\int_{0}^{\infty} e^{-s t} \frac{\sin a t}{t} d t=\arctan \frac{a}{s}$$ and $$\int_{0}^{\infty} e^{-s t} \frac{\sin a t \cos b t}{t} d t=\frac{1}{2} \arctan \frac{a+b}{s}+\frac{1}{2} \arctan \frac{a-b}{s}.$$ Identifying \(\alpha=t, x=a,\) and \(y=s,\) the solution of Problem 14 is $$\begin{aligned} u(x, y) &=\frac{100}{\pi} \int_{0}^{\infty} \frac{1-\cos \alpha}{\alpha} e^{-\alpha y} \sin \alpha x d \alpha \\ &=\frac{100}{\pi}\left[\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-\alpha y} d \alpha-\int_{0}^{\infty} \frac{\sin \alpha x \cos \alpha}{\alpha} e^{-\alpha y} d \alpha\right] \\ &=\frac{100}{\pi}\left[\arctan \frac{x}{y}-\frac{1}{2} \arctan \frac{x+1}{y}-\frac{1}{2} \arctan \frac{x-1}{y}\right]. \end{aligned}$$

Step-by-Step Solution

Answer

The solution is given by \(u(x, y) = \frac{100}{\pi}\left[\arctan \frac{x}{y} - \frac{1}{2} \arctan \frac{x+1}{y} - \frac{1}{2} \arctan \frac{x-1}{y}\right]\).

Step 1: Understanding the Problem

We are given a Laplace transform problem where we need to evaluate a specific integral involving trigonometric functions part of Laplace transforms. The problem involves using known Laplace transformations about \(\sin\) and cosine functions.

Step 2: Identify Function Structure

By observing the problem, we note that \(u(x, y)\) is expressed using an integral with terms \(1 - \cos \alpha\), \(\sin \alpha x\), and an exponential decay term \(e^{-\alpha y}\). This allows us to separate the integration into two known components of Laplace transforms.

Step 3: Separate the Integral

Recognize that \(\int_{0}^{\infty} \frac{1 - \cos \alpha}{\alpha} e^{-\alpha y} \sin \alpha x \, d\alpha\) can be split into two integrals:1. \(\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha} e^{-\alpha y} \, d\alpha\)2. \(- \int_{0}^{\infty} \frac{\sin \alpha x \cos \alpha}{\alpha} e^{-\alpha y} \, d\alpha\)

Step 4: Apply Known Laplace Transforms

Apply the given Laplace transforms:1. For \(\int_{0}^{\infty} e^{-\alpha y} \frac{\sin \alpha x}{\alpha} \, d\alpha = \arctan \frac{x}{y}\)2. For \(\int_{0}^{\infty} e^{-\alpha y} \frac{\sin \alpha x \cos \alpha}{\alpha} \, d\alpha = \frac{1}{2} \arctan \frac{x+1}{y} + \frac{1}{2} \arctan \frac{x-1}{y}\)

Step 5: Construct and Simplify Solution

Substitute the known integral solutions back into the expression for \(u(x, y)\) and simplify:\[\begin{aligned} u(x, y) &= \frac{100}{\pi} \left[\arctan \frac{x}{y} - \frac{1}{2} \arctan \frac{x+1}{y} - \frac{1}{2} \arctan \frac{x-1}{y}\right].\end{aligned}\]

Step 6: Final Expression of Solution

The transformed and evaluated integral confirms the given expression, combining results from both integrals: \(u(x, y) = \frac{100}{\pi} \left[\arctan \frac{x}{y} - \frac{1}{2} \arctan \frac{x+1}{y} - \frac{1}{2} \arctan \frac{x-1}{y}\right]\).

Key Concepts

Integral Calculus

Integral Calculus is a fundamental branch of calculus focusing on the accumulation of quantities and the areas under and between curves. It is all about finding the integrals of functions. Integration can be thought of as the reverse process of differentiation. In this exercise, integration is used to evaluate definite integrals from 0 to ∞ that appear in the expressions related to Laplace transforms. These definite integrals help us transform a function of time into a function of a complex variable, making certain calculations more manageable. Here are some key ideas that can assist with understanding:

• Definite Integrals: They have limits of integration, here from 0 to ∞, representing accumulation over an interval.
• Laplace Transforms: This is a technique to transform a complex function involving exponential growth or decay into a simpler algebraic form.

Integration in this context is crucial for handling complex differential equations and finding solutions that might be difficult otherwise.

Trigonometric Functions

Trigonometric functions like sine and cosine play an important role in Laplace transforms, especially when dealing with periodic or oscillating functions. In our problem, the sine function (\(\sin\alpha x\)) and its combination with cosine (\(\cos\alpha\)) appear prominently in the integral expressions. These functions help model phenomena such as wave forms or harmonic oscillations. Let's break down some essentials:

• Sine Function: \(\sin\) can denote oscillations, like those seen in waves.
• Cosine Function: Often paired with \(\sin\) for phase differences in wave-like behavior.

Understanding these functions allows us to visualize how they influence the behavior of the solution, hence interpreting their impact on the transformed equations.

Exponential Decay

Exponential decay is all about how quantities reduce over time, often rapidly at first, then slower. In our exercise, the term \(e^{-\alpha y}\) signifies exponential decay in the integral expression. This decay is a vital aspect when dealing with Laplace transforms as it allows for simplification of the process by diminishing effects over long time periods. Here are some things to ponder:

• Decay Factor: This is represented by \(e^{-\alpha y}\) and influences how quickly values reduce.
• Laplace Transform Role: The exponential decay term helps reframe complex time-domain functions into a s-domain that is simpler and linear.

The exponential decay is not just an abstract mathematical idea; it represents real-world phenomena like heat dissipation, radioactive decay, and more, leading to manageable mathematical representations of these scenarios.

Arctan Function

The \(\text{arctan}\), or inverse tangent function, plays a pivotal role in our final solution. It transforms ratios into angular measures. In this case, it helps express the result of our integrals in a simplified manner. The original problem integrates several arctan expressions, which results in the neat outcome we see in the final equation for \(u(x, y)\). Here's a closer look:

• Inverse Function: It helps convert values from trigonometric ratios back to angles.
• Simplification: Utilizes arc relations to express otherwise complex integral results succinctly.

Understanding the \(\text{arctan}\) function not only rounds off integration solutions but also enhances comprehension of relationships within trigonometric contexts that might be otherwise opaque.