Problem 10 - Differential Equations and Linear Algebra

Question

Use \((10.10 .1)\) to determine \(L[f]\). $$f(t)=\left\\{\begin{array}{cc} 2, & 0 \leq t \leq 1 \\ 1-t, & 1 < t \leq 2 \\ 0, & t > 2 \end{array}\right.$$

Step-by-Step Solution

Answer

The short answer is: $$ L[f](s) = -2\frac{e^{-s}}{s} + \frac{2}{s} + \frac{1}{s^2}(e^{-s} - e^{-2s}) $$

Step 1: Define each part of the function

For our piecewise function, we have: $$ f(t) = \begin{cases} 2, & 0 \leq t \leq 1 \\ 1-t, & 1 < t \leq 2 \\ 0, & t > 2 \end{cases} $$ We can rewrite this as three separate functions: $$ f_1(t) = 2, \quad 0 \leq t \leq 1 $$ $$ f_2(t) = 1-t, \quad 1 < t \leq 2 $$ $$ f_3(t) = 0, \quad t > 2 $$

Step 2: Compute the Laplace transforms of the individual functions

We will find the Laplace transforms for each part of the piecewise function using the definition of the Laplace transform: 1. For \(f_1(t)=2\), we have: $$ L[f_1](s) = \int_0^1 e^{-st} (2) dt $$ 2. For \(f_2(t)=1-t\), we have: $$ L[f_2](s) = \int_1^2 e^{-st} (1-t) dt $$ 3. For \(f_3(t)=0\), we have: $$ L[f_3](s) = \int_2^\infty e^{-st} (0) dt $$

Step 3: Solve the integrals to find the Laplace transforms

1. For \(L[f_1](s)\), we have: $$ L[f_1](s) = \int_0^1 e^{-st} (2) dt = 2 \int_0^1 e^{-st} dt = 2\left[\frac{-e^{-st}}{s}\right]_0^1 = -2\frac{e^{-s}}{s} + \frac{2}{s} $$ 2. For \(L[f_2](s)\), we have: $$ L[f_2](s) = \int_1^2 e^{-st} (1-t) dt $$ Perform integration by parts, with \(u=1-t\) and \(dv = e^{-st} dt\), we get \(du=-dt\) and \(v = -\frac{1}{s}e^{-st}\) $$ L[f_2](s) = (1-t) \left(-\frac{1}{s}e^{-st}\right)_1^2 - \int_1^2 -\frac{1}{s}e^{-st}(-1) dt = \\\ [-\frac{1}{s}(1-t)e^{-st}]_1^2 + \frac{1}{s}\int_1^2 e^{-st} dt $$ $$ L[f_2](s) = -\frac{1}{s}(e^{-2s} - 2e^{-s}) + \frac{1}{s}\left[\frac{-e^{-st}}{s}\right]_1^2 = \frac{1}{s}(2e^{-s} - e^{-2s} - e^{-s} + e^{-2s}) = \frac{1}{s^2}(e^{-s} - e^{-2s}) $$ 3. For \(L[f_3](s)\), we have: $$ L[f_3](s) = \int_2^\infty e^{-st} (0) dt = 0 $$

Step 4: Combine the Laplace transforms

Finally, we combine the Laplace transforms found in Step 3 to find the Laplace transform of the original piecewise function: $$ L[f](s) = L[f_1](s) + L[f_2](s) + L[f_3](s) $$ $$ L[f](s) = -2\frac{e^{-s}}{s} + \frac{2}{s} + \frac{1}{s^2}(e^{-s} - e^{-2s}) $$ The Laplace transform of the given piecewise function \(f(t)\) is: $$ L[f](s) = -2\frac{e^{-s}}{s} + \frac{2}{s} + \frac{1}{s^2}(e^{-s} - e^{-2s}) $$

Key Concepts

Piecewise Function

A piecewise function is a type of function that is defined by different expressions over different intervals. In the given exercise, the function \( f(t) \) is defined in three distinct pieces:

• For \( 0 \leq t \leq 1 \), the function is constant at 2.
• For \( 1 < t \leq 2 \), the function is linear, defined by \( 1-t \).
• For \( t > 2 \), the function takes a value of 0.

This indicates that the function behaves differently in each defined interval, which makes it a piecewise function. Such functions are quite common when modeling situations where a process might change behavior at different stages or periods. Understanding piecewise functions is essential because, in many real-life scenarios, system responses are not consistent and exhibit differing behaviors over different periods. This concept lays the groundwork for further topics like Laplace transforms.

Integration by Parts

Integration by parts is a technique used to solve integrals, based on the product rule of differentiation. In mathematical terms, if you have two functions, \( u \) and \( v \), the formula is:\[\int u \ dv = uv - \int v \ du\]In our exercise, during the calculation of the Laplace transform of the piece \( 1-t \):

• Set \( u = 1-t \) and \( dv = e^{-st} dt \).
• The differential \( du \) becomes \( -dt \), and \( v \) becomes \( -\frac{1}{s}e^{-st} \).
• Plugging these into the integration by parts formula simplifies finding the integral.

Using integration by parts is crucial when dealing with complicated integrands that are products of functions, as it can simplify the integral into a manageable form.

Inverse Laplace Transform

The inverse Laplace transform is used to find the original time-domain function from its Laplace-transformed counterpart. The process involves finding a function whose Laplace transform gives the transformed function you have. Inverse Laplace transforms are helpful in situations where you're given the behavior of a system in the frequency domain and need to determine how it acts over time.
While our exercise focuses more on obtaining the Laplace transform, understanding the inverse process is important. This is because many real-world applications require converting back to the time domain after analysis in the Laplace or frequency domain, such as in engineering fields for solving differential equations or control systems.

Step Function

The step function, often represented with the Heaviside function \( H(t) \), is a basic tool used to model functions that switch values at specified points. In piecewise functions, the step function can succinctly represent sudden changes in value, and it often plays a key role in transforming such functions to the Laplace domain.
In the given function \( f(t) \), each segment can be viewed as modified by a step function at a point in time. The key property of the step function is that it turns on, or steps up, from 0 to 1 at a certain point, which can help simplify the mathematical representation of a piecewise function. Understanding how step functions work is crucial because they simplify the analysis and computation of Laplace transforms for switching systems and piecewise functions.