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Chapter 6: Problem 9

Find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{array}{ll}{0,} & {t<\pi} \\ {t-\pi,} & {\pi \leq t<2 \pi} \\\ {0,} & {t \geq 2 \pi}\end{array}\right. $$

Short Answer

Expert verified
$$ f(t)=\left\\{\begin{array}{ll}{0,} & {t<\pi} \\\ {t-\pi,} & {\pi \leq t<2 \pi} \\\ {0,} & {t \geq 2 \pi}\end{array}\right. $$ Answer: The Laplace Transform of the given function is: $$ F(s) = \frac{1}{s^2}\left(e^{-2\pi s} - e^{-\pi s} s\right) $$

Step by step solution

01

Understand the function

The given function f(t) can be written as: $$ f(t)=\left\\{\begin{array}{ll}{0,} & {t<\pi} \\\ {t-\pi,} & {\pi \leq t<2 \pi} \\\ {0,} & {t \geq 2 \pi}\end{array}\right. $$ Now, we will consider each case and find the Laplace Transform for each case, which can be summed to find the Laplace Transform of the given function.
02

Case 1 (t < 蟺)

When t < 蟺, f(t) = 0. The Laplace Transform F(s) for this case will be: $$ F_{1}(s) = \int_{0}^{\pi} 0\cdot e^{-st} dt = 0 $$ So, F鈧�(s) = 0
03

Case 2 (蟺 鈮� t < 2蟺)

When 蟺 鈮� t < 2蟺, f(t) = t - 蟺. The Laplace Transform F(s) for this case is: $$ F_{2}(s) = \int_{\pi}^{2\pi} (t-\pi) e^{-st} dt $$ Now integrate by parts, where we let u = t-蟺 and dv = e^{-st} dt: $$ \begin{aligned} F_{2}(s) &= \left[-\frac{1}{s}(t-\pi)e^{-st}\right]_{\pi}^{2\pi} - \int_{\pi}^{2\pi} -\frac{1}{s} e^{-st} dt \\\ &= \frac{1}{s}(t-\pi)e^{-st}|_{\pi}^{2\pi} + \frac{1}{s^2} e^{-st}|_{\pi}^{2\pi} \end{aligned} $$ Plug in the limits of integration and simplify: $$ F_{2}(s) = \frac{1}{s^2}\left(e^{-2\pi s} - e^{-\pi s} s\right) $$
04

Case 3 (t 鈮� 2蟺)

When t 鈮� 2蟺, f(t) = 0. The Laplace Transform F(s) for this case will be: $$ F_{3}(s) = \int_{2\pi}^{\infty} 0\cdot e^{-st} dt = 0 $$ So, F鈧�(s) = 0
05

Combine results

Now we combine the results from Steps 2, 3, and 4 to find the Laplace transform of the given function: $$ F(s) = F_{1}(s) + F_{2}(s) + F_{3}(s) $$ Substitute the obtained Laplace Transforms for each case: $$ F(s) = 0 + \frac{1}{s^2}\left(e^{-2\pi s} - e^{-\pi s} s\right) + 0 = \frac{1}{s^2}\left(e^{-2\pi s} - e^{-\pi s} s\right) $$ Therefore, the Laplace Transform of the given function is: $$ F(s) = \frac{1}{s^2}\left(e^{-2\pi s} - e^{-\pi s} s\right) $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions
Understanding piecewise functions is essential when studying various mathematical concepts, including the Laplace transform. In essence, a piecewise function is defined as a function composed of multiple sub-functions, each of which applies to a specific interval in the domain. Each 'piece' of the function provides a rule for calculating the function's value within its respective interval.

For the case of the given exercise, we look at the function that changes its formula depending on the interval of the variable, in this context, time 't'. It is defined as zero before the time reaches 蟺, changes to 't - 蟺' between 蟺 and 2蟺, and reverts back to zero as time is equal to or exceeds 2蟺. This type of function allows us to model situations that are not uniform across an entire range, making them incredibly useful in real-world applications like engineering and physics.

To find the Laplace transform of such piecewise functions, we consider each 'piece' separately over its interval, which matches the approach in the given solution. By treating these intervals individually, we simplify the process of finding the Laplace transform of the entire function.
Integration by Parts
Integration by parts is a powerful technique used to solve integrals where the function to be integrated is a product of two functions. This method comes in handy, particularly when dealing with Laplace transforms where such products frequently occur.

The method is based on the product rule for differentiation and is formally expressed by the formula: \( \int u dv = uv - \int v du \), where one function is designated as 'u' and the other as 'dv'. The goal is to choose 'u' and 'dv' wisely so that the integral of 'v du' becomes simpler than the original integral.

In the solution provided, integration by parts is utilized when the Laplace transform of the function 't - 蟺' is computed. Here, 'u' is chosen to be 't - 蟺' making 'dv' equal to 'e^{-st} dt'. The resulting integrals are much simpler and can be easily evaluated, showcasing this technique's power in breaking down complex integrals into more manageable parts.
Differential Equations
Differential equations form the backbone of modeling the behavior of various systems in science and engineering. These equations involve mathematical functions and their derivatives, describing the relationship between a function and its rates of change.

In the context of the Laplace transform, differential equations often arise when analyzing systems over time. The Laplace transform is a tool that converts differential equations into algebraic equations, which are easier to solve. After solving the algebraic equation, the inverse Laplace transform is used to convert back to the time domain, providing the solution to the original differential equation.

Although the exercise provided doesn't directly involve solving a differential equation, understanding the link between Laplace transforms and differential equations is crucial. When functions are defined by differential equations, particularly for dynamic systems, the Laplace transform enables a streamlined approach to finding solutions by working with the transformed algebraic equations instead of directly tackling the differential equation itself.

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Most popular questions from this chapter

Find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related. \(y^{\mathrm{iv}}-y=u_{1}(t)-u_{2}(t) ; \quad y(0)=0, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=0, \quad y^{\prime \prime \prime}(0)=0\)

Determine whether the given integral converges or diverges. $$ \int_{0}^{\infty} t e^{-t} d t $$

For each of the following initial value problems use the results of Problem 28 to find the differential equation satisfied by \(Y(s)=\mathcal{L}[\phi(t)\\},\) where \(y=\phi(t)\) is the solution of the given initial value problem. \(\begin{array}{ll}{\text { (a) } y^{\prime \prime}-t y=0 ;} & {y(0)=1, \quad y^{\prime}(0)=0 \text { (Airy's equation) }} \\ {\text { (b) }\left(1-t^{2}\right) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0 ;} & {y(0)=0, \quad y^{\prime}(0)=1 \text { (Legendre's equation) }}\end{array}\) Note that the differential equation for \(Y(s)\) is of first order in part (a), but of second order in part (b). This is duc to the fact that \(t\) appcars at most to the first power in the equation of part (a), whereas it appears to the second power in that of part (b). This illustrates that the Laplace transform is not often useful in solving differential equations with variable coefficients, unless all the coefficients are at most linear functions of the independent variable.

Find the Laplace transform of \(f(t)=\cos a t,\) where \(a\) is a real constant.

Suppose that \(F(s)=\mathcal{L}\\{f(t)\\}\) exists for \(s>a \geq 0\) (a) Show that if \(c\) is a positive constant, then $$ \mathcal{L}\\{f(c t)\\}=\frac{1}{c} F\left(\frac{s}{c}\right), \quad s>c a $$ (b) Show that if \(k\) is a positive constant, then $$ \mathcal{L}^{-1}\\{F(k s)\\}=\frac{1}{k} f\left(\frac{t}{k}\right) $$ (c) Show that if \(a\) and \(b\) are constants with \(a>0,\) then $$ \mathcal{L}^{-1}\\{F(a s+b)\\}=\frac{1}{a} e^{-b / d} f\left(\frac{t}{a}\right) $$
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