Question

As in Example 2, use the definition to find the Laplace transform for \(f(t)\), if it exists. In each exercise, the given function \(f(t)\) is defined on the interval \(0 \leq t<\infty\). If the Laplace transform exists, give the domain of \(F(s)\). In Exercises 9-12, also sketch the graph of \(f(t)\). $$ f(t)= \begin{cases}0, & 0 \leq t<1 \\ t-1, & 1 \leq t\end{cases} $$

Short answer

#Answer# The Laplace transform of the given piecewise function is: $$ F(s) = \frac{\mathrm{e}^{-s}}{s^2} $$ with its domain being all real numbers \(s > 0\). The graph of the piecewise function \(f(t)\) consists of a horizontal line at \(y = 0\) for \(0 \leq t < 1\) and a straight line with slope 1 passing through the point \((1, 0)\) for \(1 \leq t\).

Step-by-step solution

Define the function and its domains

Our function is: $$ f(t)= \begin{cases}0, & 0 \leq t<1 \\\ t-1, & 1 \leq t\end{cases} $$

Split the Laplace Transform Integral

Since \(f(t)\) is a piecewise-defined function, we split the Laplace transform integral into two parts, one for each piece of the function: $$ F(s) = \int_0^{\infty}\mathrm{e}^{-st}f(t) \,\mathrm{d}t = \int_0^1 \mathrm{e}^{-st}(0) \,\mathrm{d}t + \int_1^{\infty}\mathrm{e}^{-st}(t-1) \,\mathrm{d}t $$

Evaluate the first integral

In this case, the function \(f(t)\) is 0 for \(0 \leq t < 1\). So, the first integral is: $$ \int_0^1 \mathrm{e}^{-st}(0) \,\mathrm{d}t = 0 $$

Evaluate the second integral

For the second part of the function, \(f(t) = t - 1\) for \(1 \leq t\). Let's evaluate the second integral: $$ \int_1^{\infty}\mathrm{e}^{-st}(t-1) \,\mathrm{d}t = \int_1^{\infty}(t-1)\mathrm{e}^{-st} \,\mathrm{d}t $$ To evaluate this integral, we can use integration by parts, with \(u = t-1\), \(du = dt\), \(dv = \mathrm{e}^{-st}dt\), and \(v = -\frac{1}{s}\mathrm{e}^{-st}\). Thus, we have: $$ \int_1^{\infty}(t-1)\mathrm{e}^{-st} \,\mathrm{d}t = -(t-1)\frac{1}{s}\mathrm{e}^{-st}\Big|_1^{\infty} +\frac{1}{s} \int_1^{\infty}\mathrm{e}^{-st}\,\mathrm{d}t \\ = -\frac{1}{s^2}\mathrm{e}^{-st}\Big|_1^{\infty} $$

Combine the integrals

Now, we combine the results of both integrals to obtain the Laplace transform: $$ F(s) = 0 -\frac{1}{s^2}\mathrm{e}^{-st}\Big|_1^{\infty} = \frac{\mathrm{e}^{-s}}{s^2} $$

Determine the domain of F(s)

The Laplace transform \(F(s)\) exists for \(s\) such that the integral converges. In this case, the function converges for all \(s > 0\): $$ \text{Domain of }F(s) = \{s \in \mathbb{R} \mid s > 0\} $$

Sketch the graph of f(t)

To sketch the graph of \(f(t)\), plot the two parts of the piecewise-defined function. The first part is a horizontal line at \(y = 0\) for \(0 \leq t < 1\). The second part is a straight line with slope 1 passing through the point \((1, 0)\) for \(1 \leq t\).

Key concepts

Piecewise Functions

Piecewise functions, such as the one in the example problem, are defined by different expressions over different intervals. This is particularly present in physics and engineering where conditions change over time or space.

For the given function \( f(t) \), it features two pieces: it is zero for \( 0 \leq t < 1 \) and then changes to \( t-1 \) for \( t \geq 1 \). These functions require special attention when we're finding their Laplace transform since the transform is performed over each interval separately and then the results are combined.

Understanding the intervals and the behaviors of functions across these intervals is crucial, ensuring that calculations are done correctly for each part.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It is useful for integrating products of functions and is expressed by the formula \( \int u dv = uv - \int v du \).

In the problem, we're applying this technique to evaluate the integral of the second piece of the function, \( \int_1^{\infty}(t-1)\mathrm{e}^{-st} \,\mathrm{d}t \), choosing \( u = t-1 \) and \( dv = \mathrm{e}^{-st}dt \). This method simplifies the integration process and is essential when dealing with functions that are products of polynomials, exponentials, logarithms, or trigonometric functions.

While it can initially seem complex, mastering integration by parts is invaluable for dealing with a wide range of integrals in calculus.

Convergence of Integrals

The convergence of integrals, particularly in the case of the Laplace transform, is about whether the integral has a finite value when taken over the indicated range. If the integral does not converge, the Laplace transform does not exist.

In our exercise, the second integral converges if the value of \( s \) is such that the exponential decay, \( \mathrm{e}^{-st} \), dominates the polynomial growth of \( t-1 \). To get a convergent result, and thus a valid Laplace transform, we need \( s > 0 \). The convergence aspect is pivotal in determining the domain of \( F(s) \), ensuring that we define the Laplace transform only where it is meaningful.

Domain of Laplace Transforms

The domain of Laplace transforms refers to the set of values of \( s \) for which the transform exists, essentially where the integral that defines the transform converges. The convergent region depends on the growth rates of the function components and the exponential factor, \( \mathrm{e}^{-st} \).

In the example problem, the domain is \( s > 0 \) indicating that the transform exists only for positive values of \( s \). This criteria is vital for the practical application of Laplace transforms in solving differential equations, as it informs us where our solutions are valid and offers insight into the stability properties of the system described by the differential equation.