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Section 4.1 The Laplace Transform

Subsection 4.1.1 Definition of the Laplace Transform

Our first transform method is the Laplace transform, which takes a function \(f:[0,\infty) \to \R\) and produces a function of a new variable \(F(s)\text{.}\)

Definition 4.1.1.

Let \(f\) be a function on \([0,\infty)\text{.}\) The Laplace transform of \(f\) is defined by the integral operator

\begin{equation*} \mathcal{L}[f] = \int_0^\infty f(t) e^{-st} \, dt = F(s). \end{equation*}

Before we get into when we can use the Laplace transform and why it is so powerful, we'll start with some examples of computing the transforms of some basic functions.

We'll start by computing the Laplace transform of \(f(t) = 1\text{.}\) Recall that when we're dealing with improper integrals (with limits at \(\infty\) in this case), we need to consider the integral as shorthand for a limit.

\begin{align*} L[1] \amp = \int_0^\infty 1 \cdot e^{-st} \, dt = \lim_{B \to \infty} \int_0^B e^{-st} \, dt\\ \amp = \lim_{B \to \infty} \left(\frac{-1}{s} e^{-st}\right) \bigg\rvert_0^B\\ \amp = \frac{-1}{s} \lim_{B\to\infty} e^{-sB} - 1\\ \amp = \frac{1}{s}. \end{align*}

So \(\mathcal{L}[1] = \frac{1}{s}\text{,}\) as long as \(s >0\) (or the improper integral fails to converge, a condition that will usually apply).

Now let's look at \(f(t) = t\text{,}\) which will require integration by parts.

\begin{align*} \mathcal{L}{t} \amp= \int_0^\infty t \cdot e^{-st} \, dt = \lim_{B \to \infty} \int_0^B t e^{-st} \, dt\\ \amp= \lim_{B \to \infty} \frac{-t}{s} e^{-st}\bigg\rvert_0^B - \int_0^B \frac{-1}{s}e^{-st}\, dt\\ \amp = \lim_{B \to \infty} \frac{-B}{s} e^{-sB} - 0 + \frac{1}{s} \lim_{B \to \infty} \int_0^B e^{-st}\, dt\\ \amp= 0 - 0 + \frac{1}{s}\mathcal{L}[1] = \frac{1}{s^2} \end{align*}

so long as \(s > 0\text{.}\) (Note that it is very easy so show that as \(B \to \infty\text{,}\) we get \(\frac{-B}{s}e^{-sB} \to 0\) by L'Hospital's rule.)

Use the same idea as the previous example to show that

\begin{equation*} \mathcal{L}[t^n] = \frac{n!}{s^{n+1}}, \hspace{1cm} s > 0 \end{equation*}

for any positive integer \(n\text{.}\)

Now consider the exponential function \(e^{at}\) for some constant \(a\text{.}\)

\begin{align*} \mathcal{L}[e^{at}] \amp= \int_0^\infty e^{at} \cdot e^{-st} \, dt = \lim_{B \to \infty} \int_0^B e^{-(s - a)t} \, dt\\ \amp = \lim_{B\to\infty} \frac{-1}{s-a} e^{-(s-a)t} \bigg\rvert_0^B\\ \amp= \frac{-1}{s - a} \lim_{B \to \infty} e^{-(s-a)B} - 1 = \frac{1}{s-a} \end{align*}

as long as \(s > a\text{.}\)

Our final introductory example concerns the basic trig function \(f(t) = \cos bt\text{.}\) (We can use parts here or just note that

\begin{equation*} \int e^{at}\cos{bt} \, dt= \frac{e^{at}}{a^2 + b^2}(a\cos bt + b \sin bt) \end{equation*}

which is a standard table integral.) Then,

\begin{align*} \mathcal{L}[\cos bt] \amp= \int_0^\infty \cos bt \cdot e^{-st} \, dt = \lim_{B \to \infty} \int_0^B e^{-st} \cos bt \, dt\\ \amp= \lim_{B \to \infty} \left[ \frac{e^{-st}}{s^2 + b^2} (-s\cos bt + b \sin bt)\right]\bigg\rvert_0^B\\ \amp= \frac{s}{s^2 + b^2} \end{align*}

Use a standard integral and the argument in the previous example to show that

\begin{equation*} \mathcal{L}[\sin bt] = \frac{b}{s^2 + b^2}, \hspace{1cm} s > 0. \end{equation*}
Notice that at this point, we have constructed the Laplace transforms of the basic forms of forcing functions that we consider in the method of undetermined coefficients (which is no accident!).

Subsection 4.1.2 Basic properties of the Laplace transform

One of the most important properties of the integral is that it acts linearly on the integrand. That is, for constants \(a,b\) and integrable functions \(f,g\text{,}\) we have

\begin{equation*} \int af + bg = a\int f + b \int g. \end{equation*}

Because the integral is linear, often operations that are defined in terms of integrals are also linear. This is the case with the Laplace transform as long as it converges, since

\begin{align*} \mathcal{L}[af(t) + bg(t)] \amp= \int_0^\infty e^{-st}(a f(t) + b g(t)) \, dt\\ \amp= a \int_0^\infty e^{-st}f(t) \, dt + b \int_0^\infty e^{-st} g(t) \, dt\\ \amp = a \mathcal{L}[f(t)] + b \mathcal{L}[g(t)] \end{align*}

This often gets recorded as the equivalent conditions

\begin{equation} \mathcal{L}[f+g] = \mathcal{L}[f] + \mathcal{L}[g]\tag{4.1.1} \end{equation}

and

\begin{equation} \mathcal{L}[cf] = c \mathcal{L}[f].\tag{4.1.2} \end{equation}

Compute the Laplace transform of \(f(t) = t^2 + 3e^{2t} + \sin 3t\text{,}\) including the domain.

Linearity makes this easy, since

\begin{align*} \mathcal{L}[2t^2 + 3e^{2t} + \sin 3t] \amp= \mathcal{L}[2t^2] + \mathcal{L}[3e^{2t}] + \mathcal{L}[\sin 3t]\\ \amp= 2\mathcal{L}[t^2] + 3\mathcal{L}[e^{2t}] + \mathcal{L}[\sin 3t]\\ \amp= 2\frac{1}{s^2} + 3\frac{1}{s-2} + \frac{3}{s^2 + 9}, \hspace{1in} s > 2. \end{align*}

Combined with the results from the previous discussion, we can now compute the transforms of a lot of the most important forcing functions that we consider when we learn constant coefficient linear equations. So far, it seems like we haven't gained much of an advantage, but the next section should make clear that the Laplace transform can handle functions far beyond what can be done with earlier methods. Also still open is the big question:

How does the Laplace transform apply to differential equations?

Subsection 4.1.3 Piecewise continuous functions

The most important quality of the Laplace transform is that it can be used on functions that are piecewise continuous. In physical situations, this correponds to forcing functions that change at points of time. For example, a continuously applied constant force could suddenly become a harmonic force modeled by a trig function at some time, and then at a later time, the force could disappear. Functions that change definitions, but which remain continuous in between those points of change are called piecewise continuous functions.

Definition 4.1.10.

A function \(f(t)\) is a piecewise continuous function on an interval \([a,b]\) if the interval can be divided up into finitely many subintervals \([a_i,a_{i+1}]\) so that

  1. \(f\) is a continuous function when restricted to each \([a_i,a_{i+1}]\text{,}\) and
  2. \(f\) does not possess a vertical asymptote at any of the points \(a_i\) (that is, \(f\) approaches a finite limit at the end of each subinterval)

If \(f\) is piecewise continuous on every interval of the form \([0,b]\) for any constant \(b\text{,}\) then \(f\) is piecewise continuous on \([0,\infty)\text{.}\)